Trigonometry AS-Level Math | Pure 1

Master the entire AS-Level Trigonometry chapter (Pure Mathematics 1) in one video! 🚀

In this lesson, I cover all the key concepts you need to know for your AS-Level exams, step by step, with worked examples and clear explanations. This is a complete guide to trigonometry in AS Pure Math 1, perfect for Cambridge, Edexcel, and other exam boards.

📌 Topics covered in this video:
– Radian measure
– Sine, cosine, and tangent functions
– Graphs of trigonometric functions
– Trigonometric identities and equations
– Solving trigonometric equations in degrees and radians

💡 What you’ll get:
– A full chapter walkthrough in simple language
– Step-by-step worked examples

Whether you’re preparing for AS-Level exams or building a strong foundation for A-Level Mathematics, this video has you covered.

📌 **Watch More Solved Papers:**
– A-Level Paper 11 (M/J 2025) ► [https://youtu.be/pHTXb-33QBk?si=qb7PheGdOmwLqX-H]
– A-Level Mechanics Paper 42 (M/J 2025) ► [https://youtu.be/-DM5ET5GZQ4?si=weSqSjkNzU8DAiI2]
– A-Level Pure Math Paper 12 (M/J 2025) ► [https://youtu.be/H2G8a7Bh1GE?si=j3RfQn8dEAGIeu5E]
– A-Level Integration Playlist ► [https://youtube.com/playlist?list=PLxImrPaCGX4am_10WqFyu4S0Oq8d6e6U8&si=iJB6jHUS3QWM3izJ]

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Yo, what’s up? Can anybody hear me? Is the sound quality good? Okay. So um as requested uh sorry as voted I put up a post about um what topic you guys want me to do from pure math one and it was you guys basically voted for um trigonometry. So that’s what we are going to be doing today. Pure one trigonometry. And yeah, let’s uh let’s start I guess. So this right here, this table that you guys can see, this is called a trick table. I’m just going to go into full screen mode here. Uh I think that’s that’s okay. I’m just going to check if this is properly visible on the stream. Uh give me a second. Yeah. Okay. I think that this is properly visible. Okay. So, what is a trig table? Well, so if if you look at the first row, we sort of have all of these um these things. I’m just going to get a pen over here. So, we have all of these guys uh these these guys. So, these are basically the angles. As you can see, they’re right in Oops. What did I do? And supposed to do that. I’m going to go back up. So all of these values these are the values for this thing. This is called this is the angle angle theta. And so you have the main possible angles that you are going to be seeing in your trigonometry pure one is going to be all of these angles. So you have 0° 30° 45° 60° and then 90°. And then of course that there are the radian versions as well which is 0 radians by 6 radians p by4 radians. Now what is a radian? Well a radian is a different way of measuring an angle. It’s just like you have degrees Celsius and degree fahhe for temperature. You have radian and degrees for measuring an angle. Now how do you go from degrees to radians? And how do you go from radians to degrees? Well, there there’s this thing called pi by 180 and then there’s this thing called 180 by pi. And so the way you do it is basically if you want to go from radians to degrees then what you do is uh so if you have let’s say 60° and you’re trying to go into radians so you basically whatever you’re trying to go into that thing has to be in the numerator. So if if I’m trying to go from degrees to radians, then then pi has to be in the numerator. Whereas let’s say if I have um pi by p by 3 and I’m trying to take this into into the degree mode, then what I’ll do is I’ll multiply this with 180 over pi. So if you’re trying to get degrees, 180 has to be in the numerator. If you’re trying to get radians, pi has to be in the numerator. That’s how you go from uh degrees to radians and radians to degrees. Now, under these uh thetas, under these possible angles, you have sin theta, cosine theta, and tan theta. And what these do is essentially see if you have all of these angles these angles theta you put these in in in the sign function you put these in the cosine function you put these in the tan function what’s going to happen is you’re basically going to get some value out of it. So that’s what the values are what we’re going to write in these boxes. So what is sin theta when when the angle is 0° or 0 radians? So that and and you have to learn this. You can use a calculator but you have to learn this like you need to know these these values off the top of your head. So I’ll just teach you guys how to construct this table and then when you go to your exam you can just construct this table at in the beginning and then you can just sort of come back to this table whenever you want to look look up a value. So sine zero. So we’re we’re so basically you put this angle in here then you’re going to put that angle in here. And so all of those angles are going to have specific values. So sin 0 is is zero. And then sin 30 or sin by 6 is 1 by 2. And then sin 45 is going to be 1 by under 2. And then sin 60 is going to be under root 3×2. And then s of 90 is going to be 1. So these are the values you need to know. You need to remember what value you get for what angle in in the case of sign. And then cosine is is basically just the flip of this. So if if you’re going from left to right now, you just go from right to left. So what I mean by that is is basically for sign for the sign you’re going from here to here. From the cosine, you’re just going to start going in reverse. And I’ll show you guys how that works. So basically cosine theta you put one over here and then you put uh for for cosine 30 you put under root 3x two then you put one by under root2 and you put 1 by two and then you put zero. So you can see how sin theta and cosine theta are going in reverse. Now tan theta is is a little is a little different than than the first two. So 10 the key values I can tell you over here the the easiest values are so 10 10 0 is going to be zero that’s straight up as simple as that 1090 is going to be infinity or undefined so you may even see or undefined undefined so infinity undefined whatever you want to call it and then 10 45 is one so these are the main values that you need to know and what about 30 and and 60 so The way you write that is see over here you have this 30 right and so it’s the 30 involves a three so then 10 tan theta is also going to involve a three what it involves is 1 over under3 that’s one way of learning it and then tan 60 is just the reciprocal of this so it would just be under root3 that is the trick table and you need to know these values and and one easy way of of doing this is just draw this table learn how to draw this table and whenever you want to pick up a value just go and look at the value from this table. That’s a really easy way of doing it. Now we basically have graphs once you’ve once you’ve sort of learned these values. You basically have these graphs. So you can basically plot out the graph of sin theta. You can plot out the graph of cosine theta. You can plot out the graph of tan theta. And what we commonly do is is see your angles go in a circle. they have a cycle. So basically if I draw a circle over here, how many degrees is this circle? So let’s say if I start over here and I go all the way around this circle, come back over here to this point. How many degrees have I covered? So I’ve covered 360°. That’s how big a circle is. How that’s how big a circle is. And in in radians, 360° corresponds to 2 pi 2 pi radians. Okay. So a cycle or a circle is is 360°. And so we basically split this into four quarters. So we’ve split it into four quarters. So this is where we’re at zero. Then this is 90° and this is 180°. Then this is 270°. And then this back is going to be 360° or 0°. It’s the same thing. Now we split it into four quarters. And then we use these quarters to sort of you know deal with the other values of s cosine and tan etc etc. Okay coming coming to these graphs. How do we sort of use this table to sort of plot out the graphs for sin theta cosine theta and tan theta and what do their graphs look like? So and and you do also need to remember the shapes of these graphs. So when I’m I’m done drawing out the shapes of these graphs you do need to remember what they look like. So what I want you guys to notice over here is that see we just talked about a cycle and we were like we’re going to split it into four quarters where this is zero this is going to be 90 this is going to be 180 this is going to be 270 and then that is going to be 360 and see that’s actually what I split it into so this is zero then you have 90 then 180 then 270 then 360 so we basically split the the graph into into those four quarters and on the x-axis we represent the theta the angle and on the y-axis we represent whatever value we get when we put that angle theta in in the sign function. Okay. So you put that angle in the sign function you’re going to get some value if you represent that value on the y-axis. So we’ve we’ve what we’ve done over here is we’ve split the cycle into four different quarters and using those quarters we’re going to plot out y= sin theta. So let’s start off with the first point z what is the value of y when when theta is equal to z. Then again you have this sort of this table right you can sort of go and check out the value of sin theta when your angle is zero and sin theta is basically zero at that point. So you put the first dot right over here. Then you go to 90° and you check out what the value of of y is at that point. So you can clearly see it’s one. You come back over here and you put one over here. And then you check out what the value is at 180°. Now, this is a little bit tricky. You’re like, well, if I go back into the table, I’m not seeing any 180°. So, how am I supposed to do this? Well, a cool thing a cool thing is that if you have this sort of this cycle, what happens is that the values that are corresponding, so let’s say 0 and 180°, they correspond to each other. and then 90 and 270 they correspond to each other. So what I mean by that is let’s say you you get um you get at s at sin 0 the answer you get is zero. So then at sin 180 you’re also going to get zero. sin 80 you’re also going to get zero. sin 180 you’re also going to get zero. Now ideally whatever number you have over here you get the negative of that over here. But see thing is zero doesn’t have a sign. So zero would just be zero. But if if I talk about this 90° this 90° what is what is sin 90? Let’s check it out. Sin 90 sin 90 is equal to 1. So this is equal to 1. Now sin 270 is also going to be one but it’s going to have a different sign right it’s going to have the negative sign and now the same thing would have applied over here if whatever if you have a positive number over here you’re going to have that same number but the negative of it on the other side problem is zero doesn’t have a sign so zero just stays at zero so you don’t know what sin 180 is but you know that sin 180 corresponds to zero so that means that sin sin 180 is also going to be zero. You put a dot over here and next point is is 270. So what is sin 270? You go into your table, you do not have a value for it. But what you do know is that 270 corresponds to 90. It’s on the opposite side of 90. So it’s going to have the same number but the opposite sign. So if sin 90 is 1, sin 270 is going to be one as well, but it’s going to have a negative sign. It’s going to be negative 1. That’s going to be somewhere around here. And what is sin 360? So remember when you complete all of this cycle, this this first starting point is also 360°. So 0 is going to be zero and 360 is also going to be zero. They’re both at the same position, right? So this is going to be zero. Okay. Now, once you’ve done this, you need to know how to draw out this curve cuz it can be something like this. like this, like this, like such. It can also be something like this, like this, and then something like this. So, what the what’s what the shape is going to be, you do have to memorize that shape. And that shape basically is going to look something like this. So, you have a curve going up like this. You have a curve coming down like this, kind of like a rainbow. And then you have a flipped rainbow. So, it would be something like this. And like this this is what a sign graph looks like. Now this is super important. This is what one cycle of of a sign looks like. So one we went one cycle consists of 360°. So we went over those 360° and we figured out that this is what one cycle of a sine wave looks like. This is really important. You do need to have this rough idea of what a sign looks like in your head. Now we go over here to the graph of cosine theta. And and this time instead of having the the angle been given in degrees, it’s given to us in radians. And we need to know how to deal with radians as well because you’re going to be dealing with both radians and degrees. We’re going to plot out the the graph of cosine theta in the same way we plotted out the graph of sin theta. So what is cosine 0? Let’s check it out. Cosine 0 is 1. So it’s going to be right over here. You put a point right over here. And then what’s cosine pi / by 2 or cosine 90°? So that’s equal to zero, right? Cosine 90 is equal to zero. So we put a point right over here. And then what’s uh cosine pi? Now pi is 180°. So how do we again remember you plot out your circle and you realize that if zero is over here then 180 is its corresponding part and at at uh cosine 0 your value is one. So when you split it to the other side when you send it to the other side it’s still going to be one. It’s just going to have a negative sign next to it. You put a dot right over here. And then what’s uh cosine 3 p by 2 which essentially means 270. So cosine 3 by 2 or 3 p by2 means 270°. You can convert it using the method I just told you guys at the start of this video and you will find out that it’s equal to 270°. So if if you have 90° over here and you have 270° over here, then what’s cosine 90? It’s basically equal to zero. So then the opposite side will just be zero as well. And zeros have no signs. So it doesn’t matter. Signs don’t matter over here. They put a dot at this point. And then what’s going to be cosine 2 pi? So 2 pi is 360. 360° is at the same position as zero. So it’s going to be one. Going to be one over here. And again this again this time you do need to memorize how the shape of the curve is going to look like because it can be something like this or it could be something like this. So it’s really important that you know how the shapes work. So you draw a curve sort of like this and then you draw a rainbow for the bottom part sort of like this and then you draw a curve like this. Okay. So this is what one cycle of a cosine graph looks like. It looks something like this. Okay? Kind of like kind of like a weird U shape or a weird we shape. So this is the rough drawing of a uh of a cosine graph. Now how do you draw a tan graph? So tans are a bit tricky. They’re not as simple as signs and cosiness, but we’re going to learn how to draw a tan graph. Now again what is the the y value at at tan 0? So basically when tan is zero so when tan 10 10 10 of 0 degrees you basically get an answer of of zero. So you put your point you put your point right over here at this point. And then what is 10 at 90°? So if you check it out 10 at 90° is actually infinity. Now how do you deal with infinities? Infinities are not common numbers. So the way you deal with them is you basically draw a line dotted line sort of like this at 90° and then you put a dash over here and you put a dash over here. That’s what you do for infinity. And and then what you basically do is you draw a dotted line wherever at whatever angle you get infinity you draw a dotted line and put the dashes like this. So the dashes aren’t like this. Don’t put the dashes like this. This is going to be wrong. Put the dashes in the pattern I show you in. Okay. And then what is what do we do for 180°? So what is and 180°? So what is 180? 180 is basically if we split it into the the four quarters again this is zero this is 180°. So if this right over here if tan 0 is zero then tan 180 is going to be zero as well and there are no signs for zero. So we don’t have to worry about signs. So we put a dot right over here. And then what about 270? So 270 is right opposite to 90°. So that means we’re going to have infinity over here as well. So if 90 was was infinity, we’re going to have infinity over here as well. So again, how the way you draw infinity is you basically you basically draw a dashed line going through whatever angle you have infinity on and you’re going to put one dash over here and you’re going to put another dash over here. And finally, what about 360? 360 lies on this point over here and it’s going to be the same as whatever zero was. So at at tan 0 the answer is zero. So at tan 360 the answer is going to be zero as well. You put a point over here. Now how do you join these points? Are are these points joining like this like this like this? This is important. How you join the points is really important as well. So the way the tan graph joins is you first draw a curve like this. All the way up to that dash and all the way up to that dash. What does that even mean? That means it’s going all the way to infinity. It doesn’t have a fixed value. Going all the way to infinity like crazy. And then you basically start from this lower dash and you bring it up all the way to this zero. And then you basically flip the direction of the curve and you send it back up. Sort of something like this. And you do the same thing once more. So you bring this curve up sort of like this. So that’s what a rough drawing of a tan graph looks like over over 360°. So this is how you draw your graphs of sin theta cossine theta and tan theta for for 360° or or 2 pi. Now moving on to the next part. There’s something you can do to basically alter the shape of these graphs. And you basically this is the rough you can say sort of a blueprint of what’s going on. So you have y equals and and see these are for sine functions cosine function tan function. So it doesn’t matter what you have the pq and r represent something really useful and I’ll tell you guys what they represent. So let’s say I take a graph of sign. So a sign looks something like this. So this is y = sin theta. And the graph looks something like this. Now you do have to remember the rough shape of this. Okay? Where this is going to be 360°, this is going to be 270°. This is going to be 180°. And then this is going to be 90°. Right? I’m just splitting it into the four quadrants we talked about. And we saw that this was going all the way up to one. This was zero over here. And then at the bottom it was going all the way to -1. So this is what the rough sketch of a sign graph looks like. Now what happens when you start adding these these constants, these these P’s, these Q’s, these Rs, what do they what do they even mean? So R is basically for the baseline and baseline is is where is a starting point where your graph starts at. So if you look at the baseline over here, it’s it’s this line. The baseline is basically wherever your graph starts from. Now this line, this line in pink, this is the baseline. This is where my sin theta starts off from. So you have this thing going up. It started from this point. It was going up all the way and now it’s going down. So, so what would be the value of R in this case? Well, the R is going to be zero because the baseline is is this is this zero line. Okay. Now, what if what if I do something different? So, what if I I had this this point over here. Let’s say I had this part over here. I draw a line on one and I and I drew my sign function over here. Something like this. So now my baseline would basically be equal to to one. First of all when I when I was dealing with the first example my baseline was zero. Now I basically started at one. I I drew my sign curve at one. So now my baseline my value of r is going to be one. So r represents the baseline. This r this r this r this represents the baseline. It’s it’s a little different for a tan function. So we’re just going to ignore tan functions for now. It’s a little different for tan functions. Tan functions are not that simple. So a little different. But for cosine and sine functions, talking strictly about cosine and sine functions, the r represents the baseline. So and what is a baseline? A baseline is where you start your um curve at. That’s what a baseline is. What about Q? What is Q? So Q is is how is I’ve even written over here. So Q is how the number of waves you complete in one cycle. So what does that mean? So let’s say one sine wave looks something like this. Now over here if you look at this this sin theta can you guess the value of Q? So Q is basically equal to one over here. And the reason for that is because well if you put a one over here then one theta is just going to be equal to theta. So what that means is it it means the number of waves you’re completing in one cycle and one cycle is of 360°. So you’re completing one sine wave in 360°. This is what one sine wave looks like. Let’s say I I change the number over here from one to two. What’s going to what’s going to be the the curve? What’s going to be this graph over here? What’s what’s the graph going to look like right now? So, it’s it’s basically it’s basically going to look like two sine waves. So, this would be one sine wave and then this would be another sine wave and you would be completing two sine waves in in one cycle or two sine waves in 360 degrees. So, it would basically look something like this. Like this. That’s one sine wave and that’s going to be another sine wave. So you’re going to be completing two sine wave in 360°. And the way I guess that is basically if if you’re going to be completing two sine waves in 360° then that means you’re going to be completing one sine wave in in in the half of 360° and the half of 360° is going to be 180°. So basically one one sine wave gets completed in 180° and then the next sine wave gets completed in the remaining angles. So you have a graph that looks like this. This is what happens when I change Q to Q to two. Now let’s say I change Q change Q to 1 by two. So I change Q to a half. Now, this means you’re going to be completing half a wave in 360 degrees. So, a sine wave looks something like this, right? Now, what what does a half sine wave look like? A half sine wave looks something like this. So, I’m going to be completing half a wave in 360° now. So, it’s going to look something like this. That’s half a wave that’s being completed in 360°. So Q dep Q Q changes the number of waves that you complete in one cycle. Q tells you how much waves you’re going to have in one cycle or how much waves how many waves you’re going to have in 360° or in 2 pi radians. That’s what Q means. So R means baseline. Baseline is where you start your curve at. Q means the number of waves you complete in one cycle. What does P mean? So P is for amplitude. What is amplitude? Amplitude is basically the number of units you go up from the baseline and the number of units you go down from the baseline. So what what do I mean by that? So in in in this example, this line zero is our baseline. Now amplitude is how much do I go up from this baseline and I go up by one unit. How much do I go down? I go down by one unit. So the amplitude the amplitude P is going to be equal to one. What if I change P to let’s say two. So right now P is one over here. But what what what happens when I change P to two. Say I change this to two. Now my my amplitude is going to be two. And I’m going to have -2 over here. This is going to change. This is going to go all the way up till -2. All the way up sorry all the way up till positive2 and then all the way up till -2. That’s what amplitude does. Amplitude is basically how much you go up from the baseline, the maximum amount that you go up from um the baseline, and the maximum amount that you go down from the baseline. That’s what amplitude means. So now we know what PQ and R mean. Let’s try out some questions to cement this idea. And we have uh this question over here from it’s taken from the past papers. And we can see that we have a graph which is y = a cosine bx + c. And see what they’ve done if is we we had p q and r. They’ve replaced them with a b and c. So that doesn’t matter, right? The num the symbols they don’t matter. We know what they’re talking about. Okay. Uh we need to find the values of of the positive integers a, b, and c. So we know what they represent. We know what a represents. We know what B represents. We know what C represents. So C represents the baseline. B represents the number of waves. So the number of waves in one cycle or or 360° which is one cycle is 360° or it’s 2 pi radians. And then a represents the amplitude. amplitude is how many blocks do we go up from the baseline and how many blocks do we go down from the baseline. Okay, so we have a cosine function over here. Now a cosine function, what does a rough cosine function looks like? It looks something like this. And if I draw out the axes over here, this is what a rough cosine function looks like. Okay. Okay. First thing I can see right off the bat is is that there are two cosine two cosine waves being completed in 2 pi. So 2 pi this is one cycle. I can see two waves being completed in one cycle. So the first wave starts from here and ends over here. Right? That’s what one waves one wave looks like. And the next wave starts over here and ends over here. So those are two waves that are being completed in one cycle. So what is B going to be? B is going to be two. B is going to be two. What about C? So C is the baseline. Now how do we figure out the baseline? This is a tricky part. How do we figure out the baseline? Now see the way you figure out a baseline is if if this is the baseline you have equal number going up and equal number going down. So if let’s say this if you say this is my baseline this is not your baseline because you’re going up less than you’re going down. They both have to be equal in number. So how do we choose a baseline? Okay. Okay. So now each block over here each block represents two two units. If you go from two to four that means you went two units. So this this comes from a bit of trial and error. But let’s say I say this is my baseline. So so I check how many blocks do I go up or how many numbers how many units do I go up. So from 4 to 8 I go up by four units. What about from 4 to -2? How many units am I going down? So, this is two units down. This is two units down. This is two units down. So, I went down by uh 1 2 3 4 5 6. I went down by six units. So, this cannot be my baseline cuz those numbers are not equal. So, how about how about I change my baseline and I put my baseline right in the middle of of two and four. So, in in the middle of two and four, you have three. Okay. Now, how much do I go down? So, this is going to be one from if you want to go from 3 to two, you’re going down one unit. And then from 2 to 0, you’re going down two units. And then you’re going down two units again. So, you go down basically by 5 units. And then how much do you go up? This is from 3 to 4, it’s one unit. Then from four to six, it’s two units. Then from 6 to 8, it’s two units. You go up by five units. So, this seems like a good baseline. So C has to be three. B has to be three. Now what about the amplitude? Well, we’ve already figured out the amplitude as well. It’s going to be five. Going to be a is going to be five. The amplitude is going to be five. That’s how much you go up. That’s how much you go down. So then this graph is basically going to be y = 5 cossine um bx. So that’s going to be 2x plus b is going to be 3. And I I want to point out to something. You can see how instead of giving you theta, they gave you x. But that doesn’t matter. They whether they give you x or whether they give you theta, it’s the same thing. So that doesn’t matter. Don’t freak out when you look at that. So this is what the the the graph or the curve is going to look like. Okay. Going down to the next uh part, we have we have a diagram that looks uh that looks like a sign and and it it is a sign. They even show us that it’s a sign. So what we’re supposed to do is we’re supposed to state the value of P, state the value of Q, state the value of R. Okay, so it’s what is R? R is a baseline. And it’s really obvious for for a sign graph. It’s really obvious. This obviously clearly looks like the baseline. Really obvious. You can just look at it. You can see, oh, this is where the sign is starting from. That has to be the baseline. So, the baseline R is going to be equal to -2. What about Q? So, Q is the number of waves, number of waves that you complete in one cycle. One cycle is basically 2 pi or 360°. So, this is 2 pi. How much waves are you completing in 2 pi? You’re completing this much wave. this much wave in 2 pi. Okay. How much is this? If you draw a sign draw a sign curve like this, how much is this? Is this is this the full wave or is this is this the half wave? And it is the half wave. That means you complete half a wave in one cycle. You complete half a wave in 2 pi. So half you’re the Q is basically going to be 1 by 2 or 0.5 which means half a wave. You complete half a wave in 2 pi. And then what is going to be P? What is going to be the amplitude? How much do you go up from the baseline? How much do you go down from the baseline? So you’re going up by let me just erase this. This is the baseline and we go up by one. From -2 to 1 you’re going up by one. Then from -1 to 0, you’re going up by one. And then you’re going up by one again. So you go 3 units up. And then you go from -2 to -3. -3 is to4 and4 to5. You’re going down by 3 units. So p is going to be equal going to be equal to 3. So your amplitude is three. Now moving on to the next part. And what we have to do is we have a curve that has the equation y = 2 + 3 sin 1x 2x for uh x is greater than or equal to zero and smaller than or equal to 4 pi. And I’ll tell you guys what this thing means in a second. And we need to find the greatest and least values of y. And we need to sketch the curve. So we need we have this equation over here. And we have to sketch out this equation. And we’re given this this sort of these possible values for x. So what is x? So in instead of theta in this question, they’ve given us x. And that doesn’t matter. That basically means that x is our theta. So x is our angle or x is our theta. And they’re saying that the theta must be between 0 to 4 pi. So that means two cycles. We have to draw this for two cycles. From 0 to 2 pi, you’re going to have one cycle. Then from 2 pi to 4 pi you’re going to have another cycle. So this is your first cycle and then this will be your second cycle. So you’re drawing two cycles. Okay. So how do we sketch this graph? First of all we have this equation y = 2 + 3 sin 1x 2x. Let’s rearrange this because if you want to sort of draw this out you need to know what these threes twos by two what all of this represents. So we know what this thing means. D sin u theta + r. We know what this thing means. Let’s try and rearrange it and get this form. So we’ll write this two over here. And now we basically have 3 sin 1 by 2x + 2. And you can see that this is going to be in the form p sin uh qx + r. And now it’s going to click automatically. You’re going to know what they’re talking about. So basically our baseline our baseline this thing is equal to two. So our baseline is going to be two. So if if I just plot out the values over or actually I’ll do that later. So the baseline baseline is equal to two. What about what about the number of cycles that we are completing? So, we’re completing half a cycle. We’re completing half a cycle in 360°. We don’t need to write this in degrees because this question is has been given to us in radians. We’re completing 1 by two cycles in in in in uh we’re completing 1x two waves in in one cycle or in 2 pi and then p is given to us as three. So this this means the amplitude this means half a wave half a wave half a wave in one cycle in one cycle and then this means the baseline. Okay. So we need to state the greatest and least values of y and we basically get the least and the greatest values. So if this is your sign right your sign curve looks something like this. Now the greatest possible value is going to be at this point and the lowest possible value is going to be at this point. The way you find those values is basically let’s say this is your baseline. Then you just go up the amount of amplitude and you go down the amount of amplitude and then you get your maximum and your minimum value. So the least and the greatest value. How do we find them? Well, the baseline is two. So you go up by three. So if you’re at two over here and you go up by three, then you’re going to get five, right? you’re going to get five over here because you add three into two. That’s going to be uh 3 4 5 and you subtract three from it. So, you’re going to get two uh one zero. Actually, I’m I’m forgetting the forgetting the what am I saying? I’m forgetting the arithmetic. So, this is basically going to be negative 1. When you subtract when you subtract three from two, you basically get negative 1. So that’s going to be your least possible value. So the greatest value is going to be five and then your least possible value is going to be -1. Okay. So what we need to do is we need to first draw our baseline. This our baseline is going to be at two. This is our baseline. This is where we’re going to start our curve at and our sine wave at. This is the point. And this is going to be pi. This is going to be two pi. So this is pi. This is 2 pi. This is 3 pi. And this is 4 pi. Okay. And then we basically we’re going to go up to a maximum value of five. That’s going to be five. And we’re going to go to a lowest possible value of negative one. Negative one. Okay. Now let’s plot out our graph. So it’s is it going to look like this? Is it going to look like this? How is it going to look like? That’s what that’s what Q tells you. So Q tells you that you are going to be completing half a wave in one cycle. Half a wave in one cycle. What is a full sine wave looks look like? It looks like this. You’re going to be completing half of this. This is half. Going to be completing half of this in 2 pi. Okay? So from 0 to 2 pi, you’re going to be completing that much. So that means that the rest of this you’re going to be completing in in in another 2 pi. So when you add another 2 pi into this 2 pi, you’re going to get 4 pi. So your your curve is going to look something something like this over over these four pi. Okay. So the way we draw it out now is we basically we basically We get this up until 2 pi, right? And then we get the lower part. We get the lower part until 4 pi. This is how we draw it out. And you have to be careful. So see my maximum value is touching that five and my minimum value is touching that negative one. So you don’t want to you don’t want to sort of do this because then you would be going over that maximum value of five. Okay. So this is how you can then use use your knowledge that you learned over here to not only figure out what graph you have but also to sketch whatever graph you have. Okay. So going on to the next part. What we need to know now is how do you solve trigonometric equations? And that’s where graphs come in handy. That’s where we’re going to use these graphs to solve these trigonometric equations. So the first equation is sin theta is equal to 0.5. And we need to and and we we’re given this range or we’re given these possible values of of the angle theta. So we’re going to go from 0° all the way to 360°. Okay. So first of all, what is what does this even mean? What does this thing even mean? So what this thing means is that you can basically have sin theta = 0.5. We’ve solved simultaneous equations, right? We all have. And and let’s say in simultaneous equation, you have a curve that’s like this x + one. And then you have a curve like this – 2x. And what you do is you basically say y = y. And you put x + 1 = x^2 – 2x. That’s how you merge two equations when you’re solving simultaneous equations. If you realize that’s what they’ve done over here. So if you reverse engineer this, you can sort of see that this could be one equation y = sin theta. This could be another equation y = 0.5. 0.5 is 1 by 2, whatever you want to call it. And then what they’ve done is is they’ve put this y= y and they’ve constructed this equation sin theta= 0.5. So what does it mean to solve two equations simultaneously? What does it mean to solve a simultaneous equation? It it means that they will have a common solution or they will have an intersection point. That if that’s what you can recall from your coordinate geometry. So that’s what they’ve done over here. They have two graphs y = sin theta y = 0.5 and they’re trying to find a common solution. So I’m basically going to draw out this this graph. So this is your theta. This is your y. Now first of all you can just go ahead and draw a rough sketch of your y= sin theta. And we know what our y= sin theta looks like. It looks something like this. Okay. where this may be 2 pi and this may be uh 3x 2. Wait, we’re we’re supposed to do this question in degrees. Remember, they’ve given you uh the the values of theta in degrees. So, we can just go ahead do this in degrees. This is 360° and this is 270°. This is 180° and then this is going to be 90°. Okay, so this is what your sign graph looks like. And then this the maximum value of a normal sign graph is one and then it’s negative one. If of course if the amplitude is one if if we’re talking about this graph right if it has one over here that just means the amplitude is going to be one. So we have we have this curve and we have this line. This is y equals 0.5. Now what does a line of y= 0.5 look like? Well, that means you’re going to have 0.5 somewhere over here and it’s going to be a horizontal line. The horizontal line is going to look something like this. You can see these uh these two graphs sort of intersect at at two points. Now, that’s what we’re trying to find. We’re trying to find the value of theta. We’re trying to find this value of theta and we’re trying to find this value of theta. That’s what it means to say sin theta is equal to 0.5. You’re basically trying to find those two intersection points. So how do we sort of do that? Now you have you have to know that these the s graph cosine graph tan graphs they’re really similar in shape. They have symmetrical properties. So basically what that means is whatever whatever you have over here, however much angle you’re going to go over here, that’s going to be the same amount of angle you’re going to go over here. So that’s that’s by you can just look at it and it makes sort of it makes some sense that okay, if this part if you have a mountain that’s sort of like this and you have two points over here and over here and they’re like they’re they’re horizontal points. There’s a line passing through them. So it’s there are hor it’s a horizontal line and there are two horizontal points and basically however much you’re going to travel over here is going to be the same amount that you’re going to have to travel from over there the other side. That’s what what’s crazy about these graphs. That’s how you can find out the values of theta. So basically the way you do that is the way you solve this out is if if you want to find the two values of theta. One thing you can do is you can make in in this equation you can make theta the subject by sending this sign over there. So it’s going to be sin inverse. Now this is going to be 0.5. And when you solve this out you’re going to realize when you solve this out sin inverse of 0.5. You’re going to realize it gives you 30°. And you’re like well that’s really nice. So and and can you guess is is this point going to be 30° or is this point going to be 30°? And it makes sense, right? Because if this is 90, then that means 30° has to be somewhere on this side. It has to be less than 90. So, of course, that means that this point right over here is going to be 30°. 30°. But you were like, “Oh, my calculator only gave me one answer. It didn’t give me another answer. I have two intersection points over here. How do I find that other intersection point?” That’s where you use this the symmetric properties. That’s where you say, “Oh, if I have to go 30 over here, this this was 0°, right? So, you move 30° to the right side, then that must mean that I must move 30° to the left side from over here in in this position.” So, and then that makes sense, right? Again, because they’re they’re symmetrical. They have they’re similar to each other. They have those symmetric properties. So, if you move 30° over here to the to the left side, then the way you can find that out is well, you started off over here. this angle was 180°. If you’re trying to move to the left side, all you can do is you can say 180°us 30° and you are going to get 150°. So that’s going to be your other answer. It’s going to be 30° and 150°. Those are going to be your two values of theta. And it makes sense, right? Because this is this is 90 to 180. And 150 lies between these two numbers. And over here this 30 makes sense because there’s this zero and there’s this 90 and 30° lies between those two angles. So the two answers are 30° and 150°. Now the calculator is only going to tell you one answer. It’s not going to tell you two answers. So you have to sort of use your brain and use these these symmetric properties to figure out that other angle as well. Now we’re going to move to the next part. And for the next part we have a cosine graph. And the question says we have to solve cosine theta is equal to 0.39 and and and the values that are allowed for for the angle theta are are from 0° all the way to 360°. So let’s again do it in the same way. The first thing we do is we sort of draw out a rough sort of a rough graph of cosine theta. This is going to be y. This is going to be um cosine cosine um oh sorry and this is going to be theta going to be theta. This is a graph of y = cossine theta. And basically what we’re going to have is this is going to be zero. This is going to be uh 90°. Again those four quarters of a circle, right? You remember 0 90 180 uh 270 and then 360. Those four quarter. So 90 180 270 360. And then a cosine graph basically looks something like this. Sort of like this. Okay. And then this the the maximum point would be 1. Then the minimum point would be negative 1. And then again this is this is when y equ= uh 1 cosine theta. Right? The the one is how much the amplitude is going to be. If this was instead two, then your maximum and minimum values would obviously be 2 and -2. Okay, but right now we’re dealing with a cosine graph, a normal cosine graph. So the maximum value is going to be one. The minimum value is going to be neg1. Now we have cossine theta being equal to 0.39. And we can see that 0.39 is going to be if you draw a line, it’s going to be somewhere over here. This could be 0.39. I don’t exactly know where it’s going to be. This is all sort of a rough sketch. That’s just how much information you need to solve out the question like this. Don’t need much information. Going to be all the way over here. Okay. So 0.39 is intersecting the cosine graph at two points over here and over here. So we need to find out the the value of theta at these two points. So it’s one of them is going to be this value. one of them is going to be this value. Now again you you see you can sort of use some some symmetry over here. If you have you have this thing and it’s like this and you see they have there’s a line going through these two points this point and this point. One thing you can do is you can see that however much angle this this part is going to be this part is going to be the same amount of angle. Okay, I hope that makes sense cuz there again they have these symmetrical properties and we can sort of use them to find out uh other values as well. So first of all we can see this this has two intersection points. So we’re going to have two answers. So what we can do is we can find out the first answer just using this equation cossine theta= 0.39 and you send the cosine to the other side. It becomes cossine inverse of 0.39. And when you solve this out you are going to get cossine inverse of 0.39. So, it gives you I’m going to round this off. This gives you 67 [Music] uh.05° 67.05°. Okay. So, where is where is the angle going to be? Is is is this angle is this angle going to be 0 point this one going to be 67.05° or is this one going to be 0 67.05°? And it makes sense. It’s going to be this one. Why? Because this is 90. So 67.05 can only be less than 90. It can’t be more than 90. So the first angle, this guy is is 67.05°. Okay. So then if if we know the first angle, how can we sort of find out that second angle? Now, now see you need to know this sort of this part because you know that this part is equal to that part. So you you found out the first angle, right? The first angle you find found out that if you go to the right this much, you’re going to get an angle which is 67.05 and that’s going to be one answer. So theta is equal to 67.05. That’s one answer. And but you can see there is another answer. You can see there is another intersection point. The way you find that is you look at this guy and you say, “Oh, this guy is going to be equal to that guy.” And and then the way you do this is you say, “Okay, well, how do I find this green part? How do I find this green part?” And you can find that by saying 90 – 67.05 going to give me that green part. So you subtract 90 subtract 67.05 05 from 90 and you get 22.95°. Okay. So now you have if this that means this part is 22.95°. [Music] Now how do I know what this blue part is going to be? How do I know what this blue part is? Well, you know this part is 270, right? You just add of 270. You just add 22.95 in there. You’re going to get So you are going to get 2 92.95° [Music] and 292.95°. There you go. Those are going to be your two angles. Now we start off with with the with the third question and it says tan x minus one equals 0. And this time you’re not allowed a full cycle. You’re allowed from 0 to pi. Pi is 180°. So you’re just allowed to to draw half a cycle. You’re not allowed to draw a full cycle. Okay? So you have 10 tan x – 1 = z. So we don’t know how to solve this. We don’t know how to solve tan n x – 1 = 0. But guess what? We do know how to solve. If if I send this thing over there to the other side, you make tan the subject. Now you do know how to solve this. It basically means that tan x is equal to one. So this will be one graph. This will be another graph and you have to find their intersection points. So basically I can draw out my tan graph over here. y = tan theta. This is my theta. This is my y ais. And I’m only allowed values up until pi in this case. So I’m not going to use anything greater than pi. I mean, you can draw out uh the graph if you want to. You can draw it up to 2 pi if you want to, but you’re just going to be wasting time. You don’t have to do that. So we have zero, we have pi by 2, and we have pi. And see, I’m not drawing it in 360°. I’m not drawing it in degrees this time. I’m not sketching out my graph in degrees. The reason for that is because this is not a degrees. This is not a degree question. They gave me these values in pi. So that means I have to do this question in in in pi. Okay. So I do remember the graph of y= tan theta and at pi by 2 which is 90° I had an infinity sort of like this. So I had this dash over there, this dash over here, and then basically what was happening was one of the graphs, one of the line curves went up like this, and then the other curve went down like this. And I don’t have to draw anything uh more than more than pi cuz that’s not necessary for this question. And then basically one the other the other graph is going to be y = 1. So I could say one could be somewhere over here and one would basically be going through this through this thing sort of like this. So I can I can see only one intersection point right there and and see if you expanded the graph all the way to 2 pi. I’m just going to do that. This is 3 pi by 2 and 2 pi. If you expanded the graph then you would have got something like this. something like this. And then this line you can see two intersection points. But see we we don’t this this is this right this pi. We need everything everything before this pi. Everything outside of this pi is is irrelevant because that’s not see the values they gave us go from 0 to pi. So you don’t need that second intersection point. Of course, it it is going to intersect at that point, but that’s just not relevant to us. So, we only have one intersection point to deal with. So, how do we sort of deal with this intersection point? I was not supposed to erase that line. How do we sort of deal with this intersection point? Okay, so this part, this part is what we’re supposed to find this angle right over here. So what we can do is we can basically say tan x is equal to 1 or tan theta is equal to 1 whatever you want to call it. In this case they call it x so we can call it x as well. Then the tan to the other side that’s going to be tan inverse of 1. And this this is actually one of those values that’s in the table. So if you go back to the table, you can see that when tan is 1, you basic when tan theta is 1, the angle you get is basically 45° or by 4. In this case, since we’re doing radians, we’re going to call it p by 4. So pi by 4. Why isn’t this going down? Thank you. Pi by 4. So x is going to be pi by 4. And of course, it’s just there’s just one sort of angle. So, it’s going to be it’s going to be this one. It’s going to be p by4. This thing is going to be p by4. And that makes sense, right? Because p by2 is going to be a bigger number. P by2 is going to be a bigger number than p by4. So, p by4 has to come before p by2. So, it and and that’s where that’s where this this point lies. It lies before pi by2. So, it’s going to be p by4. Right? If you divide one by two and you divide 1 by four, what’s the smaller number? Obviously 1×4. That’s what’s happening over here. Pi by two is the greater number. Pi by 4 is the smaller number. So we we got our answer and that’s what we needed. So this is going to be the value for x. Now we go to point now we go to question D. I’m just going to erase this real quick. And question D says that we have an equation 2 9x + 1 = 0. How do we solve this? How do we find the values of x for this? So what you can do is again you can sort of rearrange this equation. So this will basically become sinx. You send the minus to the other this one to the other side you’re going to get minus one. And you can send this two to the other side. So you’re going to get sinx = -1 / 2. So now you basically have to solve this thing. You have to solve sinx = -1 / 2. And then you can basically plot out the graph of sinx. And what are the values that are allowed? So the values that are allowed are from 0 to 3 by 2. So 3 p by 2 is going to be 270°. So we have p by 2 over here. Then we have pi. Then we have 3 p by 2. 3 p by 2. And we have 2 pi. But that’s not relevant in this case. And then we have one over here. We have negative 1 over here. And we can draw out our sign graph. So the sign graph looks something like this. And we don’t need to draw that next part. We don’t need to draw it until 2 pi. We only need the we only we only have the values up until 3 3x 2 pi. Okay. And then uh sin x is equal to this other graph which is – 1×2. So – 1 by 2 is 0.5.0.5 maybe somewhere over here. Going to go straight like this. And it’s going to intersect at this point. And of course, it’s going to go forward as well, but that’s not necessary. So, we need sort of this point over here. Now, how do you find out this point? Now, one thing that it’s it’s it’s actually this is a really good piece of advice. Whenever you have a negative graph, like in this case, this is a negative 0.5. What I would recommend you to do is find out the positive counterpart of this. So if this is negative 0.5, what I would recommend you to do is find the positive 0.5 angle and then use use the the symmetry properties to find out whatever this angle is going to be. And if that doesn’t make sense right now, I’ll just solve this for you guys. You can see how to do this. So if you put this if you try and solve this directly, then this is basically going to be sinx = 1×2. If you put this in your calculator, sin inverse of -1×2, then you’re going to get a really weird angle. So if you put this in your calculator, put 0.5, you’re going to get -30 or or you’re going to get negative by 6. If your calculator is in radian mode, you’re going to get x= by3. But you can see all of these angles, these are in in these are positive angles, right? If if you’re you’re talking about negative p by 3, that’s going to be on this side on the negative axis. We don’t have a negative axis. We we are only allowed values from 0 to 3 3x 2 pi. So how do we sort of why is the calculator giving a negative angle? We don’t have a negative angle. Why is it giving a negative theta? So that’s a problem. So what we’re going to do is you’re not you’re never going to you’re never going to sin inverse of a negative number. You never do this. So you never do this. Never do this. Instead you basically solve the this the sin inverse for its positive counterpart and then you use the symmetry properties of of these trigonometric functions to basically find out whatever the answer of of the proper answer the real answer of this thing would have been. Okay. So that’s what we’re going to do. I’ll show you guys how to do this. But you never directly solve a negative value like this. You never take the inverse, the sign or cosine or tan inverse of a negative number. You don’t do that. It’s going to give you a negative angle. And if that negative angle is, you know, not in your possible values of theta, you’re just going to end up with the wrong answer. So, you don’t want to do this. Instead, the way I would approach this question is I would draw the positive counterpart of this and I would find the angle over here. So, so look at it like this. Look at it like this. So this is a this is a sign graph. Right? Now if if I have let’s say I have 0.5 over here and I have the same line. I have the same line but just in in a negative version. I have the same line over here just the negative version. 0.5 and 0.5 are the same thing. The negative just means it’s a it’s a negative version. It’s a different version but they’re the same line. Right? What I would do is I would find out this part. So these two these two things and then these two things. So you could basically find out either this part or this part and then this part would basically be the same as whatever whatever the blue part was. That make sense? Because because they’re the same line, right? 0.5 and and negative 0.5 are the same line. So whatever this this blue gap is going to be, that’s what this red gap is going to be. And and let’s say this if this was a different line, let’s say this was0.3, right? Then of course then of course this wouldn’t have worked. Then this would have been over here. Then this would not have been the same. Then you would need a line of positive 0.3 to figure out the angles. Okay. So the way again the way we do this is we’re going to find out the positive counterpart and then we’re going to use the positive counterpart to find out uh the the negative the the the this this angle over here. So the positive counterpart going to give us these two these two things and you can find out any any of you can find out this or you can find out this point is that first of all both of these. So this is going to be equal to this. The blue part is going to be equal to the blue part and they’re both going to be equal to the purple part or the pink part. So it doesn’t matter which one you find. So what I’m going to do is I’m going to find out the blue part. I’m going to and this this angle was pi over here. This was pi. So I I what I put in my calculator is I I basically I say sinx is equal to 0.5. Now I’m using the positive counterpart remember and so I take the inverse of this take the sin inverse 0.5 and what I get is I basically get I by 6 what I do with this now is I realize that oh this blue part is going to be pi by 6 also is it is it go is this part is this blue part going to be pi by 6 or is this blue part going to be pi by six So look at it this way. This is pi, right? This is p by 2 and then this is p by 6. So this pi is the biggest number cuz it’s not being divided by anything. Then this is p by two. It’s being divided into a half. Then this is p by 6. It’s being divided into six pieces. So this is the smallest smallest part. So this can’t be pi by 6 because it’s right before pi. Okay? That’s that’s not going to happen. Before you’re going to get before pi, you’re going to get p by2 first and then you’re going to get p by 6. So that means this angle has to be p by 6. Okay? So this angle this this space this space that you just figured out this space is equal to by 6. This space is equal to by 6. So basically this angle that you move to the right this angle is going to be I zoom in over here this angle is going to be pi by 6. Now if you want to find out this angle what you would have to do is you would have to go in this direction by p by 6. So you take this pi and you subtract from it p by 6. you subtract from it p by 6 and you’re going to get the angle at this position. But that’s not what I’m trying to do. What I’m trying to do is I saw that this thing this gap was p by 6. This gap was p by 6. And therefore this gap is also going to be p by 6. So what I’m going to do now is I’m basically going I I already know my angle over here. This angle is pi. I need to find out this angle right over here. What I have to do is I have to go to the right by by 6. What I’m going to do is I’m going to take my pi. I’m going to add my p by 6 into it. I’m going to get some new value. Let me just find that value out. So pi + by 6 that’s going to be 7 by 6. Now that is going to be my answer for this. So my x is going to be this is this is incorrect. This is not correct. This is going to be 7 pi by 6. Okay, that’s how you do uh a question where you have where you have something going in the negative part. You basically change it into a positive counterpart. You solve for the positive counterpart and then you basically uh think about what the negative counterpart is going to be using the symmetry properties of these s cosine and tans. Now part E says we have 4 sin^ 2 x + 2 = 3 and we’re allowed the values of of x to be from 0 to 2 pi. So this is going to look something like if I solve it out it’s going to look something like um 4 sin 2x + 2 = 3. Now what what we do is we make sin x the subject first because that’s what we did in all of the previous problems. So we are going to make sin x the subject. This going to be 4 sin^ 2 x. Take this two to to the other side. You’re going to get 1. And then take this four to the other side. So we’re going to get sin^ 2 x = 1 by 4. And send the square to the other side. So we’re going to get sinx. Now see sin x has become the subject. You’re going to get under root plus minus under root of 1×4. You’re going to get two equations. sin x is equal to positive 1×2 and sinx = 1×2. So again we can sort of draw out our sign curve. So this is going to be the x. This is going to be the y. This is a graph of y= sin theta or sinx whatever you want to call it. And then we’re going from zero all the way to 2 pi. So you got zero, you got pi by 2, which is 90. Then you got um pi which is 180. Then you got 3 p by 2 which is 270. And then you got 2 pi which is 360. And then sign goes from 1 all the way to -1. 1 to 1. And then the graph looks something like this. Like this. Like this. and like this. Okay. And now we have so we have to solve it for positive 1 by 2 and forgative – 1 by 2. So I’m not going to draw out two different graphs because this is sign this is sign. So the sign graph is the same but I I can just draw out these two lines. So 1×2 I have to solve it for 1x 2 and negative 1×2. So 1x two basically means 0.5. So, I’ll have a positive 0.5 going like so. I’ll have a So, this is a positive 0.5 and I’ll have a negative 0.5 going like so. So, this is negative 0.5. And you can see they’re actually already counterparts of each other, right? These things are counterparts of each other. Positive 5 and negative 0.5 are the counterparts of each other. So if you can find the the positive values, you can use them to find the negative values because you don’t find the negative values on their own. You’re going to use the positive counterpart to find out the negative values. That’s what I’m going to do. So you’re going to get two intersection points over here, this guy and this guy. And then you’re going to get two intersection points over here. this guy and this guy and and see this this this much angle is going to be equal to this much angle and then this gap is going to be equal to this gap. So all of these gaps are equal. The reason is because these lines are 0.5. This is also 0.5. The only different difference is it’s it’s a negative 0.5 but it’s 0.5 still. So all of these these angles are equal to one another. So if you can just find one of them, you will be able to find all of them. And I’ll show you guys how to do that. So the first thing we do is obviously you’re not going to use this again. You’re not going to use this because you do not sign inverse a negative value that that gives you messed up answers. So we can go with this and we can say x is equal to sin inverse of 1 by 2 and you’re going to get by 6 and and this is actually in your table as well. If you see you have sin theta being equal to 1x 2. If you inverse this you’re going to get p by 6 or 30°. This is in your table as well. So pi by 6 that means that means this part this part is going to be p by 6. Now how do I know this is pi by 6? Well because see if this is p by 2 right p by 2 then p by 6 is less than p by 2. So it has to come before it has to come before this guy. That’s why the first angle is p by 6. Okay. Now if if so that means that means I went to the right by by 6. I went to the right by by 6. So then what’s this angle going to be? What’s this angle going to be? What’s this angle going to be? And I can find those angles out. So basically this this angle basically means I’m going to the left by by 6. And then this this this angle this guy was pi right? So if I subtract I subtract p by 6 from pi then I’ll basically know what that second angle is going to be. So p – by 6 is going to give me 5 by 6. So this is going to be 5 by 6. So these are the the blue values. The blue values are p by 6 and 5 by 6. And then the green values that I’m going to get are going to be are going to be so this this part is to the right of pi. So I just say pi + by 6. I believe we did this in the last part and this was equal to 7i by 6. So that’s going to be one value 7 by 6. And then this this angle is to the right to the left of 2 pi right this thing is 2 pi. Now I have to go to the left of 2 pi. So that means 2 pi minus / by 6 is going to be let’s solve that out. So 2i oops 2 what am I doing? 2i – by 6 that’s going to be 11 by 6 11 by 6. So that’s going to be 11 by 6. Those are going to be all of the answers for this question. Okay. So, heading on to part F. And part F says I’m going to delete this. Part F says that we have sin x= 0.5. And this time we have to go from -3 from – 360 all the way to 90°. Okay. This is the first time we’re going to involve a negative angle. So I’m going to draw out my sign graph this. So this is going to be my sign graph. We’re doing this one. Yes. So sine graph y = sinx. So we have to go all the way to 90° and then we have to go all the way to 360°. in in the negative side. So if if this is zero, then this is going to be 90 and then 90° and then – 180° and then -270° and then -3°. Okay. And on the positive side, we only have to go all the way up till positive 90°. So let’s let’s just draw this over here. Positive 90°. And then sign goes from 1 to 1. 1 to 1. Okay. So we have sine of x. It’s equal to 0.5. Let’s draw the graph of sine of x. So this is what a normal uh s graph looks like. And up up until 90° it’s it’s this this is 90°. Then this is 180. Then this is 270. Then this is 360. So we just have to draw this this part out. This part is going to look something like this. You could draw more if you wanted to, but that’s not necessary because you’re only asked to go up until 90°. And then how do we draw the negative part of this? So this is what a sign graph looks like. Right? Now imagine if see this if if I just draw out the remaining part. This is what it’s going to look like. So if if you’re if you’re to draw a graph on the negative side, how is it going to look like? Or actually, you know what? I’m not even going to draw these these sides for now. I’m just going to draw a rough sketch to make it simpler. If you draw out an angle a sign to the negative side, what is it going to look like? Is it going to look like this? This isn’t a sign, right? You you just drew this. This is not a sign curve. A sign curve looks like this. So the way you you can draw this is you can basically sort of imagine that you pick up this curve and you pick this curve and you put it over here and that is what the sign looks like. So it it it just keeps on repeating, right? If if you draw over here, it’s going to look like it’s going to look like this. The sign just keeps on going and going and going. So if you are to draw the the negative side, you are to draw the negative side of a sign graph, then you can imagine this is what a sign looks like. If I just pick it up and I put it over here, what is it going to look like? So it’s going to look something like that. Okay. And then this is obviously going to be uh negative. Wait, I have to sort of increase the height over here. It has to go to negative one and then it has to go to be all the way to negative 1 and then it’s going to go all the way to positive one and then it’s going to come down like this. Okay. So, this is going to be 90°. This is – 180°. This is -270°. This is -3°. Okay. Now this these are the possible values of x that I can have. Now I have to draw the line y equals 0.5. 0.5 can be somewhere over here. So the line is going to look something like so. And it’s going to intersect this point this point. This point and of course the line is going to go forward. It is going to intersect that point. But that’s not within our range. Our range is between this and this. So we don’t have to be concerned about what’s outside of our range. We can even erase this part. This is of no concern to us. So we’re trying to find out what this angle is, what this angle is, what this angle is. And again, we’re going to use symmetries over here. So this thing looks the same as that thing. and and this this thing also looks the same as as the other two things. So if you can just find out one of these this this gap you can find one of this gap then you can find all of the angles and I’ll show you guys how to do that. So what you do is you basically the first the first angle you can find by just saying sine of x equals 0.5 and then you take the inverse sin inverse of 0.5 and sin inverse of 0.5 is going to give you if you’re doing this in radian modes going to give you p by 6 by 6 okay so that means first of all what angle is going to be p by 6 which of these angles is going to be p by 6 so it it It can’t be the negative angle. Of course, right? Can it? This p by 6 doesn’t have a negative sign. It has a positive sign. So, it only makes Oh, actually, we we have to do this in degrees. This was a degree question. This is going to be 30°. So, 30° isn’t a negative number. So, it’s not going to be on the negative side. It’s going to be on the positive side. And it’s it’s less than 90. So, it has to be on the left side of 90. So it just makes sense that oh there’s only one intersection point that’s on the left side of that’s on the left side of 90° and is also positive and that is this intersection point. So that means this angle is going to be 30°. So that means I I went to this side by 30°. I’ll go to this side by 30° and then I’ll go to this side by 30°. So one of the angles is 30. Now, what’s going to be the second angle? So, see this this right here is 180° and you’re going to the left of 180° by by 30. So, you’re going to the left of 180° by 30. So, what you can do, you can say -1 180 minus 30 and that’s going to be – 180 – 30 going to be -210. That’s one of the angle -210. And then the other angle, see this is -3. This angle is -3 right over here. And you’re going to the right by 30°. So you can say this is going to be – 360° plus 30°. This is going to be 330°. That’s another angle. And you can see you had three intersection points and you found out this was a negative. You found out all the three angles over here. And then two of them were negative. So two of your angles are negative. One of them was positive. So one of your angle is positive as well. Okay. Now we are going to go over to part delete this. We are going to go over to part G. Part G says cossine x + 1x 2 is equal to zero. And we’re allowed to go from zero all the way to 360. So cossine x = 1 by sorry cosine x + 1x 2 = z. So we can make cosine x the subject. So this is going to be cosine x = 1×2. And then we can now sort of plot out our graph of cosine x uh from 0° to 360°. This is zero. We’re going to have 90 uh 90 180 270 360. I’m going to do that in just a second. First, what I’m going to do is I’m just going to draw out this cosine graph. And so this is going to be 360. This is going to be 270. This is going to be 180. This is going to be 90. And then this is one. And then this is negative one. Okay. Uh 1 by 1 by 2 that is going to be over here. Y is equal to – 1 by 2. 1×2 means 0.5 going to be over here. These are two intersection points that I’m going to get this guy and this guy. Now what did I say about having negative curves or negative lines? Now you don’t you don’t make the mistake of of solving this as it is and doing this. You do not do this. Put a big cross on this. You don’t do that. Instead, you go for the positive counterpart. So instead, you would maybe try out a line which is going to be positive 0.5. And so that line is going to do something like this. Going to intersect over here, intersect over here. So you’re trying to find out whatever this gap is or whatever that gap is and it’s going to be the same gap as as this gap and as this gap those are just you know symmetric properties. So we can go ahead we can say okay what is cosine x = 1. So we’re taking the positive counterpart now and we’re going to apply the inverse on this positive counterpart going to be positive 1×2. So you’re going to get uh cosine inverse of 1×2 is going to give you 60°. This is again given in your table. A lot of these values are given in your table. So cosine 1 by 2 gives you 60°. Okay. So now that we have 60° that basically means that so which of these is going to be 60°? Look, if this thing was 90 degrees, then it makes sense that 60 is going to be less than that. So, it’s this is the intersection point that is less than 90. So, this intersection point has to be 60°. So, that means you went over here by 60°. But that’s not what you’re concerned about. You’re trying to find this gap, this gap right over here, because that gap is then going to be this gap as well as this gap. So, how do we find that gap? We just say 90 – 60 is equal to 30°. And so this gap is now 30° and this gap is 30° and this gap is also 30°. So how do we find out this this second intersection point? How do we find out this intersection point? Well, you can see this angle was 90° and you went to the right by 30°. So what’s going to be your angle? So 90 + 30 is going to be be I believe 120. So one of your angles is going to be 120. This angle is 120°. And then what is this intersection point going to be? So see this is 270 right? 270. And you’re going to the left by 30°. So when you do that, you’re going to subtract uh 30° from 270° and you’re going to end up at 240°. So that’s another one of your angles. So we found out that x was basically equal to 120° and it was equal to 40°. That’s done. Okay. Now we are going to do part H. And part H is where things start to get a little bit complicated. So part H says we have cossine^ squar x and we have sin x and this is equal to 1 by 2 sin x and we these are the possible values of x that are or or the possible values of theta that are allowed. So how do we sort of solve this question? This has not only does it have cosiness but it also have it also has signs. So the first thing you want to do is you want to get rid of of whatever trigonometric function you can get rid of cuz you cannot solve two trigonometric functions at once, right? All of these other questions that we did. What we did was we basically got one trig function like let’s say sign and we got some value for it. So that’s what we’re trying to do. And one thing I can see right off the bat is I can get rid of the sign functions. So if if you send this sign to the other side, what’s going to happen is you’re going to get is equal to 1 by 2 sin x over sinx. Then these sinx cancel out. But you have to make sure of one thing before you can do this step. And that is that see this sinx can have a range of values that depends on whatever this value of x was given to you. So what the value of x that was given to you was between 0 to 90° 0 to 90° and therefore sinx can can be between whatever whatever the the sin x is going to be. So sinx is what? So if if you put zero in here if you put you put zero in here. So this is sin x right sin x if you put zero in here you are going to get um sin 0 is what sin 0 is zero and then what is sin 90. So sin 90 is one. So the value of the values of sin x can be between 0 to one. I go up over here. Let me just check if sin zero. I’m not remembering this correctly. Okay. S 0 is 0 and sin 90 is one. So basically sinx can be anything between 0 and one but it cannot be 0 and one. It can be anything between those. And the reason for that is because if you look at at at this guy this allowed values of x to be greater than zero and smaller than 90°. So sin x can be greater than zero and smaller than one. anything between these two values. It cannot be zero but and it cannot be one because this is not equal to this is just greater than and smaller than they don’t have an equal to sign. But what if they did have an equal to sign. Now you cannot cancel out those two sin x’s. The reason for that is because then you your sinx can also be zero. And so if you say sinx over sinx and let’s say they are they can be equal to zero then what you’re also saying is you’re saying that zero is is basically over zero and then you cannot cancel this out. This this doesn’t give you a valid answer. If you try to divide 0 by zero you know what you get? You don’t get an answer that’s that’s a math error that’s undefined. So you can’t do this. So over here the the entire reason I was able to cancel out the sinx with the sinx is because it could not be equal to zero. It could be anything that’s greater than zero but it could not be zero. So you have to check when you’re doing this. You have to check if sinx can be zero or if it cannot be zero. If it can be zero then you can’t do anything about it. You can’t cancel it. In this case it it was we were able to sort of cancel it because it’s not equal to zero. Okay. So we canceled out the sin x with the sin x. So we’re just going to be left with one because anything divided by itself is 1. This is going to be cosine^ 2 x = 1 by 2. And then we can under root it. So this is going to be cossine x = + – under 1 by 2. And you can you can then expand this. You can say cosine x= under roo 1 under2. And then obviously the plus minus. So under root 1 is just going to be 1. You’re going to get cossine x = pos1 by under 2 and cosine x cosine x 1 by under 2. That’s those are the two values that you’re going to get. Okay. So we have to solve this. We have to solve these two equations now. And we have to find the possible values of x. So what I’m going to do is I’m going to plot out my cosine graph. This is what my cosine graph is. This is going to be my x. This is going to be my y. y = cosine x. And what are the possible values of of the angles that I can get? So those are from 0° all the way to 90°. This is zero and then this is going to be 90 over here. And I don’t have to plot anything else. So if this, see, if this is what a cosine graph is, then you’re only allowed to draw it up until 90. So this is basically going to be 0°. This is going to be 90. This is going to be 180, 270, and then 2 pi. So I am only allowed to draw until over here. After that, nothing is of my concern. So this is going to look something like this. And this is obviously going to be one. This is going to be negative one. Okay. So, good thing is this equation is not going to give you any answers. So, you can say no answers. No answers. And the reason for that is because if this is negative one by under root2, so it’s a negative number. It can be somewhere over here, somewhere over here. That doesn’t matter. That’s not the point. Point is that this this line, let’s say it’s over here. It’s going straight. There is no cosine graph that it can intersect with because we’re only allowed from 0 to 90° and from 0 to 90° the cosine graph is only in the positive part. It’s not in this this negative part. So no matter wherever this line is because the this line is negative it’s never going to intersect with the cosine graph. Now, of course, if we were allowed more values like values greater than 90°, then the cosine graph would be doing something like this and then we would have intersection points. But right now, we don’t have any intersection points. So, we can just say we have no answers from this part. Okay. What about this other part? So, 1 by 2. First of all, let me just check what 1 by 2 is equal to. So, it’s equal to 0.707. So, that’s going to be somewhere over here. 0.707. This is positive 1 by 2. It’s going to be going like this. So we can see one intersection point over here. So what we can do is we can say x is equal to cossine inverse 1 by under 2 and this is going to be we can take cosine inverse of 1 by under roo2 and we actually have this value in our table as well. So cosine we have this this is this is 1 by under 2 when you inverse this you get 45° or p by4 we are looking for 45° or p by4 45° or p by4. So we’re dealing with in with this graph in degrees. This is going to be 45 45 degrees. And it makes sense, right? because 45° is less than 90°. So it makes sense that this thing is going to be 45. So this angle is of 45°. So we can just say x is equal to 45°. That’s the only one answer we’re going to get. And and that’s done. Okay. So moving on to part I. Part I says uh we have 3 sin 2x + 1 = 0. And then we’re given some range for 4x. Now this question is a little tricky. And the reason for that is because instead of having x or instead of having theta, we now have 2x. So how do we do such questions? Before I do this, I’m just going to check if my laptop still has battery. Okay, so we’re still going good. Okay, so we can I think we will be able to complete this in time. Okay, how do we solve this? How? Because if you go and look at the previous questions, this was x, this was x, this was x, this was x, x, x, x, theta, theta. Now, we have 2x. So, how do we deal with this? So, one thing you can do right off the bat is I I’ll tell you guys how you deal with questions like this, but first of all, let’s make sign the subject, right? So right now we have 3 sin 2x + 1. 3 sin 2x + 1 = 0. So you make sign the subject. This is going to be sin 2x = -1 over 3. Okay. Now see they they’ve given this this is 2x and then they’ve given you the range which is pi greater than x and smaller than pi. Okay, this is the range. This these are the possible values for x. Now, you don’t want to you don’t want to solve for x, do you? What do you have? You have 2x. So, what you’re going to do is you’re going to convert this in this range into 2x because you can’t sol you this range is is not what you have. So, you have to make these two the same. So the way you do that is you basically take this thing and you can multiply this entire thing by two and and if you do that you’ll realize now this turn if you multiply this with two it it does turn into 2x that’s what you have but then you have to multiply this with two and you also have to multiply this with two this is going to be negative 2 pi and that is going to be 2 pi. Now you have 2x there and you have Oops. You have Wait, wait, wait. You have What have I done? Hold on. Okay, so you have 2x there. Wait, this my pen just stopped working for no reason. Okay, so this is 2x. This is 2x. And that is 2x. So they’re now the same. Now the other thing you do is you basically give this a name. So you can call it theta. And so you give this a name and you can call it theta. So again what do you do when you have let’s say you have sine let’s say 4x minus one in here and let’s say you have the range x is greater than -1 and x is smaller than sorry x is greater than pi and x is smaller than pi. What do you do when you have a situation like this? So you can see that this range is not the angle you have. You have 4x – 1. So what you’re going to do is you’re going to try to make this thing the same as what you have. So you’re going to multiply this with four and you’re going to subtract one from it. So all of this is going to look like 4 – 1 and this is going to be 4x – 1 and this is going to be 4 pi minus one. That’s what you’re going to do. Now you are the second step is to give this thing a name. So you can call this theta and you can call this theta as well or you can call it anything else. You can call it like let’s say alpha. You can call this alpha. Anything you like doesn’t matter. Important thing is step one whenever you don’t have whenever you you’re not just dealing with sinx and you’re dealing with something else. First thing you do is you check the range. See if if the range that you have is similar to what’s in inside your s or inside your cosine or inside your tangent. Second thing you do is you give whatever you have once once you make these two things similar you give them a name. Okay. And that’s that’s going to be really useful. So once we’ve done this we’ve done these two steps now we can actually go on to attempting a question like this. Okay. So now we have we have sin sin theta = -1 by 3 and you have and now you basically have the range for theta as well right you you have a range for theta which is going to be uh theta is less than 2 pi but is greater than -2 pi also what what is what is theta theta is basically theta is basically Theta is basically 2x right you give it a name and we gave it the name as theta. So we we gave 2x a name and that name is theta. So now this is what we have and this looks way similar to what we’ve solved in the previous questions. So what we did was we took a hard question and we made it we we did some steps to make it similar to what we were doing before. And so now it’s really easy. All you have to do is you have to plot out that sign graph. We have to plot out y= sin theta. That’s going to be theta over here. y over here. So sin theta looks something like this. And then this angle. Oops. This Oh, actually we do have a negative part. So we do have to take into account the negative part as well. Okay. So we need a negative part. Well, so this could be our y and then this could be our negative part. So a sine graph would look like this and a sine graph would also look like. So if you don’t know how to draw the negative part, you can sort of imagine it in your head that a sine wave looks like this. I’m supposed to pick it up and I could put it over here and that does the trick. But if if you don’t have a problem with that, you can just sort of draw out the sign graph. Oops. Um, I have to draw this again. You can draw it again like this. Okay. So, this is zero. This is going to be 90. This is going Wait, we’re not 90. We’re We’re supposed to do this in radians. So, 90 is p by 2. This is going to be pi by 2. This is going to be pi. This is going to be 3 pi by 2. This is going to be 2 pi. And then over here we’re going to go in the negative pattern. So this is going to be negative by 2, pi, -3 p by 2 and -2 pi. Okay. So we have positive values for theta and we have negative values for theta. And now what we have to do is we have to say we have to put this equal to this line which is 1×3. Sogative – 1 by 3 is going to be 0333 going to be somewhere over here somewhere over here going to do something like this. So this is almost equal to0.3333. Okay. What do we have to do? We cannot solve negative numbers as they are. we we don’t we don’t just take sin inverse of a negative number. Instead, what we do is we try and solve its positive counterpart first. You can see this would yield four solutions. Two over here, two over here. And since they are all sort of symmetric, what you can see is that this part be the same as this part. And then this gap or this part would be the same. So all of these gaps would be the same. That’s the that’s the special part about these trigonometric functions and and sort sort of there their spe speciality you can call it. But we are not going to solve out the negative part. We’re going to convert this into the positive part. So we’re going to convert this into positive 1×3. So positive 1×3 would be 0.333 and it would be somewhere around here. as such. So this is the positive counterpart. These two intersection points and these two intersection points and all of these are symmetric. So this thing, this gap, uh this gap, this gap and then the green gaps, all of these gaps are symmetric. That’s really important. So if you just you can find out one of these gaps, you can sort of go ahead and find out all of these gaps. So that’s what I’m going to do. So remember right now we’re not solving we’re not going to be solving for the negative counterpart. We’re going to be solving for the positive counterpart. So if we say sin theta is equal to positive 1×3 and then we find the value of theta by taking sin inverse 1 by 3 we’re basically going to get sin inverse of 1 by 3 and you’re supposed to do this in radians right because this question was given to you in radians you turn your calculator into radian mode and you uh take the inverse of 1 by 3. So you get 0.3 3 98 radians. Okay. So what we just we just f we just solved this for the blue part. What angle did we get? Did we get this angle, this intersection point? Did we get this intersection point? This intersection point or this intersection point? That’s the first question. Which intersection point did we get? So it’s first of all see this radians this this answer that you got it’s is it a negative answer or is it a positive answer. So it’s not a negative answer right. So that means you can’t have this angle or this angle because these are in the negative regions. Your theta is not negative. You got a positive value for theta. Okay. Now it’s so that leaves this angle and this angle. So which angle is it going to be? Is it going to be the the the angle that is the further away or the angle angle that is closer to zero? It’s it’s really easy to answer that because see there this is pi by two. This is this is this thing is pi. So pi is basically equal to 3.142. Pi by 2 is equal is going to be the half of that. So if this is 3.142 this is going to be 1.5 something something something right? that the half of three is 1.5 and then the half of 1.42 is going to be something something and and your angle theta is is 0.3398. So is it going to be before by 2 or is it going to be before pi? So obviously it’s going to be before p by 2, right? So that means it’s closer to zero. That means the angle you just the intersection point you just found out is is this intersection point. This intersection point right over here. So, this intersection point that you just found out. I’m just going to zoom in onto this. Oops. I uh wait, my stream, something’s up with my stream is lower than the recommended. Uh let me just hold on, guys. Give me a second. We recommend to use Okay, I think this is okay. I don’t think anything’s wrong with the stream right now. Huh? Okay, the connection’s good again. Uh, okay. Anyways, back to what we were doing. Okay. Um, hold on. View full page. Right. So if I zoom in on this part, what you can sort of see is that this is my sign graph. This is the angle over here. And this angle is 0.3398 [Music] radians. And this is my this is my curve, right? So that means I went to the right by 0.3398. So what I found out is I found out this gap. this gap is equal to 0.3398. Now once I have this gap I can find out all of the other gaps. So this remember this is not your answer as of now. This is not your answer. This is just the gap that we found. Now all of the stuff that I’m going to find in green that’s going to be your answer. So this thing this thing this thing and this thing. So what I’m So this I can call this one. I can call this two. I can call this three. I can call this four. So let’s let’s start off with one. So theta is going to be so you can see this is pi. This thing is negative pi. And you are this gap to to the right of negative pi. So the way you find it is you basically if you’re going to the right you are just going to add your 0.3398 or you’re going to add the gap into your negative pi. So you’re going to get some angle which is going to be p + 0.3398. So that’s going to be -2.8 radians. Okay, that’s your first angle. Your second angle is going to be so this this was zero. This thing is 0° and you’re going to the left. So you’re if you’re going to the left by 0.3398, then that’s just going to be negative 0.3398 radians. And then for part three, uh you can you can sort of see that that this this thing is pi and you’re going to the right of pi by uh 0.3398 radians. So you can say pi plus 0.3398 That’s going to be I + 0.3398. That’s going to be 3.48 radians. And then for the last part, part four, you can see that this is uh 2 pi and you’re going to the left of 2 pi. So you can say 2i – 0.3398 and that’s going to be 2 – 0.3398. So that’s just going to be 5.94 [Music] radians. So all these are all of the the green angles that you were supposed to find. And and they make sense because see two of these things, this guy and this guy were in in the negative part. So this is a negative angle. This is a negative angle. And then this thing and sorry this thing and and this thing were in the positive part. So these two angles are positive. So that’s that’s it for your part I. Now part J. Let’s try and solve part J. So we’re going to try and solve part J. Now part J is a quadratic equation. I don’t know if you can tell or not, but this is a quadratic equation. So, tan^ 2 theta + tan theta is equal to zero. Now, you may not be able to figure it out right now, but let’s say you say uh tan theta is equal to x. So, now you can replace tan theta with x and you can replace tan^ 2 theta with x^2. And now it’s a quadratic equation. And it’s in the form ax^2 + b x + c. So c is obviously going to be zero. b is going to be 1. A is going to be one. So you basically have a quadratic equation right now. So how do we solve this quadratic equation? So you can use the quadratic formula. You can use the quadratic formula or you can factor it out. So if if you factor it out, this is just going to be I mean you can do whatever you want. You can use quadratic formula, factorization, vertex form, whatever you want. Whatever you use to solve a quadratic. I’m just going to factor this out. This is going to be, if I factor out an x, I’m basically going to get uh x + 1 = 0. I’m going to get two answers from this. It’s going to be x= 0 and x=1. So, these are the two answers that I get. Now, remember x is not what I’m trying to solve for. I’m trying to solve for tan theta. So I can replace the x with with tan theta= 0 and tan theta=1. And now I I basically have to solve these two equations. So I can plot out a graph of tan theta and I’m allowed values from from 0 to 360°. So if I if I draw out my curve, this is my y. This is my my theta. And this is going to be a graph of y equals my pen stopped working again. Okay, this is going to be y = tan data. Okay, so tan the graph of tan looks something like this. Then it looks something like this. And it looks something like this. And we have a line as such over here. I mean, you don’t have to draw this dotted line, but it’s helpful when you’re sketching the curve, but you it’s not necessary. This dotted line, you don’t have to draw it out. Okay. Uh this is 0°, this is 90°, this is 180°, this is 270°, and then this is 360°. Right? This is the graph of tan. Okay? So one I have I have a line which is equal to y = 0 and I have a line y = minus1. So y equals 0 is pretty straightforward. It’s just going to be this line y = 0 is is the line of of of the x-axis or the line of the horizontal line over here. Okay, so this is the first line and I have to find out the intersection points on this line and this is pretty straightforward. This is not difficult at all. You can just see all of the intersection points. And so this the origin of course would be your first intersection point. And then you can see this the graph is passing through 180°. This is an intersection point. And then the 360 is also an intersection point. But if you check out the range, see you can your theta your theta is allowed to be zero cuz it can be equal to zero. But it cannot be equal to 360°. So the final the this this third point is not going to be an intersection point. This is not going I’m going to put a cross on this. This is not going to be an intersection point. It’s outside of the range. So So theta is equal to uh 0° and it’s also equal to 180°. Okay, that was pretty straightforward. That was not difficult at all. Now what about the second equation? So the second equation says tan theta is equal to minus1. So that would be some place over here. Let’s say minus1 is over here. The y = -1. The line of y= -1 would be something like this. Now we know that whenever we have uh a negative value, we tend to first find out their positive counterpart because cuz that’s just simpler. So we can take y= pos1 and the line would look something like this. So we’re trying to solve for this line. Okay. And so we will have two intersection points over here. This intersection point and we will have this intersection point over here. Okay. So they will correspond to this intersection point and they will also correspond to this intersection point. Now tan is tangent is a bit tricky. Tangent is not as simple as as the rest of the problems. Tangent is a little bit tricky cuz it’s it’s not just straightforward symmetric like that. And if I draw this out a little better, I think I may be able to do a better job. So, this going to be like that. And then the intersection point. And then it’s going to go up like this. And then it’s going to go it’s going to go down like this. 360°. So that’s that’s the point over there. And that’s the point over there. Okay. So, this is going over here. This is going over here. Okay. Now, what are we going to look at? So, you can see this gap, this blue gap is going to be equal to this gap. How did I figure that out? Well, if if you see this this thing is this is the vertical part, right? And then and then for for for the this bottom part, this right here is the vertical part. So if if you’re looking at at this blue part, which is which goes from the intersection point to the vertical part, then over here, that’s what you’re going to do. This is the intersection point and you’re going to go to the vertical part, right? because you don’t want to say this thing is equal to this thing. That would be wrong. This this is this doesn’t include the vertical part. This is when you’re going when it’s when it’s over over here at the bottom and it’s and it’s curving to the intersection point. So this part would actually be equal to this part from the intersection point where it was it was sort of low like this and it went all the way to to the intersection point. So it these graphs are symmetric but you have to choose the right points. You have to choose the right gaps. You can’t just sort of sort of say whatever gap is equal to whatever gap it has to match and and this matches. So this is the intersection point and you’re going all the way to the vertical part. And then this is the intersection point and you’re going all the way to the vertical part. So this gap must be equal to this gap. Okay. So what we are going to have is is this gap and it’s equal to this gap. Okay. Now I think we can solve this. And then also this this this gap will also be equal to this gap over here. and this gap over here. So the gaps you have to sort of check out the symmetry for the gaps. You cannot not every gap is equal to every gap. The gaps have to match. And in s and cosine graphs that’s a bit easy to do. In tan graphs you have to be a little bit careful about this. Okay. So all of these blue gaps are equal to each other. So what I’m going to do is if I can find even one single one of these gaps, I can find all of the other gaps. So I’m I’m going since I took the positive counterpart, I’m going to say tan theta is equal to positive 1. Take tan inverse of positive 1 and you are going to get 45°. This is equal to 45°. So that means so which which gap which which of these angles is going to be 45? Is it going to be this one? This one? this one or so actually it can’t be the bottom part because I I I took the positive value I took the positive counterpart so is it going to be this first angle or is it going to be the second angle and it’s going to be the first angle because see if this is 90° if this line represents 90° then 45° has to be behind this 90° so that means this angle right here is 45° but I’m not trying to find that angle I’m trying to find this gap and this gap exists s between 45° and 90°. So, how do I find out that gap? I just subtract 45 from 90 and I get 45°. So, this blue gap is actually 45°. That means this gap is 45° as well as this gap is 45°. And now what I have is essentially if I raise all of this stuff, I raise all of this line because I don’t need that anymore. I basically have basically have uh this. Yes, I I have I have this. Okay. So now I know what this gap is between the intersection point and this line and then between this intersection point and this line. So these are 45° gaps. 45° 45°. So, if I’m trying to find the first intersection point, what I will do is see this this line, this line is 90° and I’m going to the right cuz the intersection point is to the right of this. So, I’m going to the right by 45°. So, you’re just going to do 90 + 45 and that’s going to give you the first angle. So, 90 + 45 is going to be I think that’s 135. 90 + 45, that’s 135. And then over here this this thing is 270 and you’re going to the right. So this is the intersection point. You’re going to the right by 45. So 270 + 45 is going to give you 315. The other angle is going to be 315. So the two answers you get from this are theta is equal to 35 and 135°. And remember this angle you found out. This was not the answer. This was just that gap that you found out. This is not the angle. This we did we did the positive counterpart just to found find out what the gap was equal to. These are the angles. These guys are the angles. So th this is basically going to be your answer. And I this this took a little bit of time. The last part took a little bit of time, but that’s just because I’m explaining it to you guys and you guys are probably doing this for the first time. The more you practice this, the easier it’s going to get and the easier you’re going to be able to see the symmetries and the easier you’re going to be able to do do these questions. So, that’s a really important thing. Practice these as much as you can. Okay. Now, moving on to the last part of trigonometry for pure one. That’s going to be your trigonometric identities. And these are the two identities that you need to know. So cosine^ 2 theta is equal to sin^ square theta and so sorry cosine^ square + sin^ square theta is equal to 1 and then sin theta / cosine theta is equal to tan theta. Now these two identities you can rearrange them. So maybe you can change this into cosine^ 2= 1 – sin^ 2 theta. Or if you make sin^ square the the subject then you’re going to have sin^ square theta is equal to 1 – cosine^ 2 theta. So those those changes you need to know that that’s also a possibility that you can do. And then over here you can also change this guy. You could take you could say sin theta is equal to cossine theta and theta. And of course uh you need to know a few identities from your O levels especially this one which is uh a² – b^ 2 is equal to a + b a minus b. I’ve seen this identity being used up many times. You need to know this identity as well. Okay. So how do you prove identities? So they they they’ll give you some fractions, some form and you will have to change it into that other form. And usually how they do this is you have the left hand side and you want to turn it into the right hand side. I wouldn’t recommend anyone to start from the right hand side. I would always say start from the leftand side and try to convert it into the right hand side. So there are some general tips from it and and the tips are number one get a rough idea get a rough idea from the question get a rough idea and and then just just solve just solve just solve and then then the second thing don’t don’t worry about about planning about planning you do not plan trigonometric trigonometric identity questions out. If you start planning, you’re going to get cooked. So I even as of now, I don’t know how this left hand part is going to convert into the right hand part. That’s not what I’m going to do. I’m just going to get a rough idea of what I’m supposed to do and then I’m just going to solve. I’m going to get the answer. The answer is going to make itself obvious. Okay? You don’t have to worry about that. So the first thing is to get a rough idea. The rough idea is I’m I’m supposed to get rid of all the signs and cosiness because I’m going to convert this uh left hand side to the right hand side. And you can see on the right hand side we don’t have any signs or cosiness. So the first thing is get rid of sin theta and cosine theta. Turn turn it turn it into nan theta. That’s the first idea, right? Second idea. I only have one fraction. I don’t need two fractions. I need one fraction. Now I can start solving this. And I don’t know how it’s going to convert into 1 / tan theta, but it’s going to happen. Trust me. If you don’t worry about it, follow the steps, it happens automatically. And you have to practice a lot of these identity questions because that’s the only way to improve at these. So we can sort of write this thing and this thing in in an easy way. So we can say sin theta uh or or one thing we can do. So we’re supposed to make one fraction, right? We’re supposed to get this fraction. We’re supposed to we now right now we have two fractions. So one thing we can do is we can take LCM. So that’s that’s the obvious thing to do is take LCM. And so 1 – cossine theta is going to be multiplied with sin theta. So we’re going to have sin theta over here minus 1 * 1 – cosine theta and then we have 1 – cosine theta * sin theta. Now I would never recommend you to expand out these denominators. You want to expand out the numerators definitely go for it. Do not expand the denominator unless it’s really required to because at the end they they usually cancel out with the numerator or something. So this is a pro tip. Don’t cancel don’t expand the denominators unless you really need to. So don’t expand denominators unless you really have to. You really have to. Okay. So then what do you do? Expand the numerators. So sin theta * sin theta. That’s going to be sin^ square theta. And then you’re going to get this. This minus one is going to multiply inside the bracket. It’s going to be minus one and plus cosine cosine theta. Okay. And then this is going to be over 1 – cossine theta time sin theta. Okay. And then what we’re going to get is see it’s what’s the next thing to do? What’s the obvious thing to do? You can apply an identity on this guy. Apply an identity on this guy. How do I know that? Well, because as I said these this identity can be converted into 9^ 2 theta = 1 – cosine^ 2 theta. Okay. So I convert this into the identity 1 – cosine^ 2 theta. This is going to be – 1 + cossine theta over 1 – cossine theta * sin theta. And then this thing cancels out this thing. And so I’m left with I’m left with minus I’m left with cosine theta minus cossine^ 2 theta over 1 – cossine theta * sin theta and and see this is why I told you not to expand the denominator because see what we’re trying to get is 1 / tan theta right and if if this is if sin theta over cosine theta is tan theta. Then then basically what’s what’s 1 / tan theta? Basically if you reciprocal this this is going to be cossine theta over sin theta is equal to and then the tan theta is going to get reciprocal as well going to be 1 / tan theta. So that’s what you’re actually looking for, right? And you can actually find that because you have see what what you can do is you can get that cosine theta and sin theta fraction. The way you do that is you can see is you already have a sin theta present in the denominator. You all you need is that cosine theta in the numerator and I can actually do that if I take cosine common from these two things. And what that’s going to do is that’s going to give me this bracket in the numerator and I’ll cancel out those two brackets. What I mean by that, I’ll just do that real quick. That’s going to be cossine theta and I’m going to get 1 – cossine theta over 1 – cosine theta sin theta. So this thing cancels out with this thing. and then sin theta over cossine theta that’s just going to be 1 / n theta. So that’s your identity that that has been proven. Now the second part again we’re going to use the same steps. Don’t freak out. Get a rough idea and and just try to solve it. So we’re trying to get rid of signs over here. So get rid of signs and um we are trying to get one fraction. As you can see in the answer we don’t have a sign but over here we have we start off with signs. So simple thing is just get rid of the signs. So uh what we can do is we can first of all take LCM to get one fraction. So, it’s going to be 1 + sin x times 1 + sin x over cossine x 1 + sin x plus uh cosine x time cossine x. And this is going to be cossine x 1 + sin x. Right? So this is going to be 1 + sin x². That’s one way I can write this cuz this is a bracket. This is the bracket. They’re both being multiplied with each other. And they’re the same thing. I mean, if you want, you can expand all of these things as well, but it’s it’s easier to do it this way. It’s up to you how you want to do this. And the denominator is going to be cosine x * 1 + sin x. Do not expand the denominator unless you really have to remember that rule. And then this is going to be cossine^ 2 x. Okay. So what do we do now? Well, we’re supposed to get rid of the signs. Um let’s expand this bracket out. Let’s expand this bracket out. And we can also expand this out. But first let’s expand the bracket out. So this is uh a + b^ 2. This would be a square. So a square is going to be 1 + b². That’s going to be sin^ 2 x – 2 a. So – 2 a is 1, b is sin x. a + cossine 2x over cossine x 1 + sin x. Okay, so we’re trying to get rid of the signs, right? We were trying to get rid of the signs. One way we can get rid of the signs is is if we open this thing up. If we open this thing up, we are going to get we are basically going to get cossine^ 2 theta = 1 – sin^ 2 theta. And that sin^ square theta can basically help us cancel out this sin^ square theta. So if we expand the cosine out, we get 1 + sin^ 2 x. And I’m I’m just going to write the numerator because I’m not doing anything with the denominator. So I’m just it’s it’s tedious work. So I’m not going to write it. This is going to be – 2 sin x and cosine^ 2 x expands out to 1 – sin^ 2 x. So this sin^ 2 x cancels out that thing. And then this one adds into this one and it becomes 2. The numerator is now going to be 2 – 2 sin x. So my fraction is going to be 2 – 2 sin x over uh cossine x 1 + sin x and then see my I was trying to convert this into 2 over cosine x already have a cosine x in the denominator and I have twos in the numerator. I don’t want this thing. So one thing I can do is I can take both of these two out as common and then I will get this this bracket in the numerator and what I’ll do is I’ll just um cancel out these these uh these two things. So basically if I take that two out as common in the numerator I will get one minus wait I I’m not did I do something wrong? I think I did something wrong in this question. Yeah, I definitely did something wrong, but I don’t remember what I did wrong. Oh, uh, this is what I did wrong. So, when I expand this thing, this is a, this thing expands into a squ + b 2 + 2 a. So, this this is a two. This two has a positive sign. This thing has to have a positive sign. I’ll have a positive sign over here and a positive sign over here and a positive sign over here. This is going to be 2 + sin x and I will get cossine x 1 + sin x. Okay, so the sin x gets canceled out with sinx and this 2 cosine x is what’s going to remain. And you can see this is what I was trying to find out. Okay, so moving on to part three. We have sin x tan x over 1 – cosine x. And we’re supposed to get rid of we’re supposed to get rid of sine and tan. So get rid get rid of them. Okay. Uh, one way I can write this fraction to make it a little bit easier for me is I could, so I’m trying to get rid of the tan, right? One way I could get rid of the tan x is I could just I know tan x is equal to sin sinx over cosine x from the identity. So this is going to be sin x / cosine x and and then it’s going to be 1 – cosine x. So I’m I’m going to turn this into a different form cuz dealing with this many fractions is a bit difficult. I’m going to change this into sinx * sinx over cossine x. And this thing is basically being divided by 1 – cossine x. Okay. So uh what we can do is we can solve this first part out. I guess this will be sin^ 2 x over cossine x / 1 – cossine x / 1 – cosine x. So what we can do is see we were trying to get rid of s and tan. We already got rid of our sign. Oh sorry we already got rid of tan. We had the tan initially. We got rid of tan by doing this. And now we we have this thing and we can get rid of sign because we know that if we use the identity uh then we can get rid we can get rid of s and we can get cosiness. So the sin^ 2 x is basically going to be equal to cossine^ 2 x 1 – cosine^ 2 x. So now we’ve got rid of that sin x as well. So we’ve got rid of all ss all cosiness sorry all s and all tans. This is just going to be 1 – cosine^ 2 x over cosine x and then divided by 1 – cosine x. And then this division can change into multiplication. So the division changes into multiplication. So that’s going to be 1 over 1 – cossine x. And so what we can do is we can um what do we have to turn this into? So we have to turn it into 1 + 1 / cossine x 1 over okay how do we do that? Uh okay. So you know what? I’m not going to expand the denominator. I’m going to write the denominator as uh 1 minus cossine x. And then I’m going to write cossine x over here. Okay. And I have to turn this thing into 1 + 1 / cosine x. 1 / cossine x. How can I do that? Interesting. I can so I need a one and you get a one basically whenever a fraction if if you have anything divided by anything or let’s say a divided by a or b divided by b. So anything divided by itself gives you one. That’s the first thing I need to know. Uh okay. So how do I solve this out? You know that identity I talked about a square minus b^ square. I can apply that to the numerator. So the way I do that and see that’s why I’m not expanding the denominator because there is still one thing I can do and until and unless I’ve done that I’m not going to expand the denominator. You want to expand the denominator as a last resort. And and the way I do that is basically I can write 1 as 1 squared cuz they’re the same thing. Right? So then this is basically going to be a square minus b square. And then I can write this as a + b and a minus b. This is going to be 1 + cos x 1 – cos x and I’ll get 1 – cos x and cos x over here. You can see this thing cancels out with this thing and then I’m going to get uh 1 + cos x over cos x and okay I I can write these separately. So if I write these separately, I’m going to get 1 / cos x plus cos x over cos x. And this is going to cancel out to give me 1. And I will basically get 1 / cos x + 1. And that’s what they actually have. So they have 1 + 1 / cos x. That’s what I have. I can write this as 1 + 1 / cos x. So I trig identities not something to worry about. Just stay calm. Don’t try to plan it beforehand. Um remember your identities and just get a rough idea from the start of the question and then just keep on solving it, solving, solving it. Eventually you are going to get the answer. Okay. So that is it for the as uh trigonometry uh uh stuff. This is going to be the pure one. all the trigonometry you need for pure one hopefully. And yeah, I hope this this live stream helped you guys and I will see you in in the next video, I