This video reviews key trigonometric identities and proof techniques for Cambridge O Level Additional Mathematics, emphasizing that only two fundamental identities are needed: sin²θ + cos²θ = 1 and tanθ = sinθ/cosθ. The instructor demonstrates how to prove various trigonometric identities by strategically applying these core identities, including compound angle formulas, double angle formulas, and fraction manipulation techniques. The session covers multiple exam questions from different years, showing how to approach complex trigonometric proofs by breaking them down into simpler components using the fundamental identities.
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2026 05 14 - Add Maths F5 - Trigonometry | Trigs PPQ review - RECAPAdded:
The whole paper is random stuff like so we never went.
>> Yeah.
>> So >> that's weird.
>> Nobody got nobody got distinctions like me.
>> Wow.
>> I can see why.
Mhm.
Yeah, I must No, no, sir. I don't do P marketing advice, right, but I not sure to tell you.
Yeah, I wish I didn't, but math home math was particularly interesting to be honest.
There's a camera talking about an idea.
Right. So, right. So, CC adm proofs and identities.
One of the most interesting parts of the syllabus.
There are only two proofs you need two identities you need to know for this to work. First one is sin^ square + cos^²= 1 and the next one is tan is equal to s cos right on the >> no it have those coming up but for the bas so I have like the compound angle ones here and I have the um yeah so you'll see double angle ones there right but so you're correct you have other things you need to know right but the two the primary identities are those two I just show you there right those kind of start everything off right now with sin^ square plus cos^ squ equal to 1 people forget you could kind of rearrange them what I mean by that is that for example if you wanted to make let's say um cos square the sub hold on yeah so for example some people like if they see cos 2 plus i^ 2 their brain doesn't compute because of course as we know addition is commutative right so where they have sin square plus cos square or cos square + sin square. The answer is still going to be one. All right.
Right. That's equal to. And then of course you could make each of these the subject in turn. If for example you wanted to make cos squared the subject all you'd have to do is subtract sin square from both sides. So you end up with 1 minus sin square theta. Similarly if you wanted to make sin squar the subject you'd subtract cos square from both sides.
And that will be that. Yeah. All right.
Cool. One sec. Cool. Nope. Sorry. Right.
All right. Now, down here, I drew a a right angle triangle and I put in a single angle theta. Right. If you draw a line from the vertex behind the angle, a straight line, whatever side it hits is the opposite side. The side opposite the right angle is always the hypotenuse.
And the third set is the adjacent side.
And we have our three basic trig ratios.
S is equal to opposite over hypotenuse.
Cosine is adjacent over hypotenuse and tan is opposite over adjacent. Right?
Now if you consider that right angles are inevitably linked with Pythagoras theorem. And what does Pythagoras theorem tell us? It tells us that the square on the hypotenuse is the sum of the squares on the other two sides.
Right? So we'll just call them opposite and adjacent because that's in the kind of context here. Right? Now if we were to do something not necessarily random but something very strategic which is we divided everything on both sides by hypotenuse squared. Sorry one second.
Uhhuh.
Well J say you join it just now right. Um yeah. So if we take opposite squared, so we dividing everything throughout the equation by hypotenuse squared, right? So opposite squared over hypotenuse squared is the same as opposite over hypotenuse all squared.
And similarly adjacent squar over hypotenuse squar is adjacent / hypotenuse and squar right and of course hypotenuse squ / itself.
You could also express like this right now based on our ratios opposite over hypotenuse is s not so opposite over hypotenuse squared will give us sin squ adjacent over hypotenuse is cosine. So adjacent over hypotenuse all squared will give us cos square. And of course hypotenuse divide by itself is one and 1 square is just going to give us one.
Yeah. So that's one way to derive it.
There are other ways as well. Right.
And for right so for tan right so tan tan theta being equal to sin / cosine right. So sin theta again is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Right? So what we're going to do is this. We're going to take your s ratio opposite over hypotenuse and divide by the cosine ratio adjacent to hypotenuse. So of course we're dividing fractions. What's the protocol for when you're dividing fractions?
Change divide to multiply and invert the divisor. Right?
What will happen is hypotenuse will cancel out and what you're left with opposite over adjacent which is right. So we have two identities that are the core this whole initial part of it. Yes the double angle and the compound angle formula will come into play later on. All right.
So actually give me a sec. I forgot I want to print out something here. Right.
Why that's so big?
Hey, here we go. Right.
Right. So, from here, cool. Print.
going to print out for you. So you have like a little booklet and what I'll do is I I'll staple that page. So you have actually Yeah, if you want you can keep it separate. So like a formula sheet, you know, if you want. So you don't have to.
So I kind of went backwards, right? The 2025 paper did have something. It had a double angle lighter. All right.
Someone passenger be able to handle the car.
What are you going to do?
Okay.
So, yeah, you can open to the to the first one. Um, you turn a couple of pages to Yeah, it should be there. Yeah.
Okay. So, we have to prove the identity that tan All right. Yeah.
Right.
Wait, what happened there? I'm the man Philip.
Yeah. So, prove that tan a + 1 / tan a is equal to 1 / sin a cos a. Okay, cool.
Um, like I said, we have two identities. We have sin square plus cos square is one and it's different um permutations and we have tan is equal to tan / cos. So which of those you think would be useful if we starting with the left hand side here?
Where is this?
It's not square. Huh?
>> So this side right?
>> Okay. What ratio you seeing on that side as in what which which trick it you seeing? Sign coine or tangent >> right. So which of the two identities would be probably more so there?
>> Okay. If you had to choose between using this identity or this one. Yeah.
>> Second one, right? Okay.
>> We said science first.
>> All right. That's okay. All right. Cool.
So, we'll see. Okay. We had to prove that. Okay. So, proof taking left hand side. So, this is some statement I was taught to write. So, that's just how I score, right? So, the left hand side is tan a + 1 over tan a, right?
Okay. So we're going to we're going to replace tan a with what? What expression?
>> Sin a over cos a. Good. And the same thing going to happen with the fraction with 1 / tan a right. So you're going to get sin a over cos a.
Okay. And now with the fraction. So it's one over the fraction which is one divided by the fraction. And again when you're dividing fractions was the protocol change divide to multiplying and invert the divisor. So you get 1 multiply by cos over s which of course is just going to give us cos over s. Now we have sin a over cos a plus cos a over sin a.
All right.
So now we have fractions that we have to add. To add fractions together, we need a common what? And the easiest easiest way to get a common denominator is to multiply, right? So cos by sign will just give us cosine or sine cause. No matter which way you put it mean the same thing, right?
So cos a by sin a.
All right. Now what we do to fix the numerators, you multiply each numerator by the opposite denominator. Right? So the sin a will be multiplied by sin a and cos a by cos a. So what's sin a by sin a sin square. And what's cos a by cos a cos.
>> Ahu.
>> And what's sin square plus cos square?
>> So we have 1 over and well cos a sin a is the same as sin a cos a. Right.
>> Yeah.
>> So if you just write it in that particular sequence to kind of match what they ask us about. So now do we have what they asked us to show? Yep.
>> Right. So now when you finish your proof you could either put well how I was taught you put a little square to show hey are done or you could put the abbreviation q d right the joy meaning is is um quite easily done the Latin is quad error demonstrandom which is what was required was show you as to you ask to show this I did it right so qed is just the symbol hey I did it I reach where you want me to reach all right how you feel about that first one there >> right okay so turn the page let's check out 2023 actually turn again. Let's do 2019.
All right. Let me just color off show that.
All right. So they have tan^ 2 + 1 is equal to 1 / cos^ 2. Now the funny part about it is that if if we if in the syllabus you had what we call the inverse ratios like one over s one over cos that kind of thing this would actually be one of the identities you'd have to kind of know off hand or be able to derive very easily. But since the inverted ratios are parallel not on the syllabus. So this right okay cool. So again on the left hand side what is the only trig item trig ratio you're seeing there tan right tan squared right. Okay so what is the identity for tan? tan is equal to correct well yes thanks right so let's start there so you so go ahead and and start off with that right so the proof we'll start as follows right taking left hand side right tan^ 2 theta + 1 Right?
So we said tan^ squ would be equal to sin^ squ over cos^ square. Right?
So we have sin square over cos square + 1. Now you could put that one over one.
So now we have fractions. And what we saw just have to add fractions. What we need a common what?
>> Um and all we have to do is multiply the two denominators. What's co^ squ by one?
Uhhuh. There you go.
And now to quote unquote fix the numerators, we multiply each numerator by the opposite denominator. Right? So the sin square will multiplied by 1, which gives us sin squ. And then the one will be multiplied by cos square, which gives us cos square. And guess what sin square + cos square is? And we end up with one over.
And that's finish it. There you go.
Mhm. How you feeling there? So far so good. All right.
Turn again to 2018 now. Right. So you So hold hold that there. I'm going to pull up 2023.
Watch what you have there for 2018 and watch 2023.
>> They repeat a question five years later.
So 2018, right? Okay, cool. So here we have more fractions, right? So but the fractions are a little more involved than the previous two questions. Would you agree?
>> All right. But what's the operation connecting the two fractions?
Subtraction. So to subtract two fractions, we need a common denominator, right? Okay. So the easiest thing to do is to multiply both of them together.
And then in the numerator, you're going to multiply each numerator by the opposite denominator, right? Okay. So start off there.
So, Right? So if you want, you could start off with your denominator looking like the two denom the two of them being multiplied like one pair of brackets by the next pair of brackets. But we will notice something down there, right?
Okay. And of course each numerator being multiplied by the opposite denominator.
So 1 by 1 + sin x will just give us back 1 + sin x. But then you minus in 1 by 1 - sin x, which of course is just 1 - sin x. All right, we up to that point.
Okay, so let's talk about the numerator, right? So when you drop the brackets in the numerator, right? You get minus one, but what will happen to the minus sign?
>> It'll change to plus.
All right? And did you spot how we could simplify the denominator?
You have a a minus b by a plus b, which is difference of two squares. So you'll get a^ 2us b^ 2 or 1 - sin^ 2 All right. Now, in the numerator, if you want, you could rearrange and put the ones together, put the sin x's together, or you could just, right? So, what's 1 - 1?
>> And what's sin x + sin x?
>> Sorry. So, we have two sin x and of course in your denominator, sorry, what is 1 - sin^ 2?
Cos 2. Right?
Okay. Now look at where we had to go, right? We had end up with 2 tan x over cos, right? Now we have 2 sin x over cos^ squ. So I'm going to show you a little technique here called decomposing the fraction, right?
Sorry, who's that? One second.
So if we have so you don't have to write it like how I did it. I just this is how I how I think through things, right?
So to so when you think of decomposition think what science and biology and things breaking down like right but that's what's going on here we breaking it up into its components. So your numerator any any aspect of the numerator could be written as by itself over one right and of course between the two and the sin x what what is what operation is understood multiplication right and the sinx is also an aspect of the numerator so you can put it over one right and of course cos square and then we have 1 over cos square right and of course we're going to break up cos^ squ into cos by cos so you have 1 / cos by 1 / cos right now tan what is the identity for tan s over cos and when you break up stuff into into it into its constituent parts, we could put together different pieces, right? So, we could put together these two. So, sin x over 1 by 1 / cos x will give us what?
>> Uh-huh. Yep. Sin x over cos x.
And guess what? Sin x over cos x is tan x. So we have 2 over 1 by tan x over 1 by 1 / cos x. And when you multiply your fraction, you multiply the numerators together and then you multiply the denominators together. Um cards cards run things. Sir I boy cards boy I holding on. Exam season in full swing. Men sleepy.
I am men.
Anyhow well thanks for asking boy. How we are you in you good? Right. And have we arrived at what they wanted us to arrive at?
>> Yeah.
>> Right. So that's why that's the purpose of going through it here so I could show you the technique. So you now you say, "Hey, I could do this." Nice. Let me do that. I'm going try something. There's a tool for your arsenal.
Why am I unprotected?
Restart.
Why why is it doing this?
Want that to happen right now? But okay, no problem. Do your thing. All right.
Um, one second.
Okay. Well, forget that. That is right.
Let's go back again. One more question.
Well, well, yeah, go back to your question. So, you have that one there.
All right. So we have 1us cos^ square over 1 + sin and that's identical to sin theta. So that's interesting right this we just kind of break down to sin theta by itself. All right cool. So again as you can see we have fractions right? Isn't that interesting? In the first one we did we had fractions and 2023 2019 2018 2017 all fractions so far right even 2014 2012 all fractions right? So technically we use more fractions than trigs here, right? So that's what's happened. The trig processes get a bad rep because of the working involved, not so much the identities, but that if you miss if you use them incorrectly, that could also throw things off, right? So sometimes, as I say, less is more. So sometimes more restraint, right? Cuz for example, a lot of times, like I said, people forget that sin square plus cos square is one. They think it only works one way. And when they see one, it doesn't occur to them. A that's sin square plus cos square. But sometimes people go they over cororrect. So like here they'll put this as s plus cos square. I'm like um you doing and then with a sign cube like we going sine cube backtrack, right? So you had to use it judiciously.
All right. Like here, right? A sin square plus cos square here might not really be of any assistance, but you never know. It might. How I would approach this is um I would still use that same what you call it >> identity right so if you look at what we have here is just sign right okay cool now cos squared let's let's let's attack cos^ squ cos^ squ is equal to 1 - sin^ square all right so proof All right, taking the left hand side.
All right, so we have 1 minus. So cos^ squ is 1 - sin^ 2 theta and that's all over 1 + sin theta. Right? Now remember in the question we worked just now we had a difference of two squares business the 1 - sin squ and we got that because we multiplied 1 - s by 1 + sign ah so yeah so you'll factoriize so you'll express 1 - sin^ square as 1 + sin by 1 - s and guess But the denominator is is 1 plus sign. So what's going to happen?
It's going to cancel with piece of the numerator.
What's going to be left?
Yeah. 1 minus the 1 - sin theta, right?
And when you drop the brackets, guess what happens? You get 1 - 1 but >> plus sin theta. And of course, 1 - 1 is zero, which leaves us with >> sin theta.
Mhm.
Right. 24. So now it starts to get a little more interesting. Right. But how you feel with the one so far?
>> All right. Cool. Right. So there is a little twist in this one. I will show you. But let's just say you encounter one. Hey J, you're going out. Thanks.
Yeah. Let's just say you encounter something like this and you didn't see the twist. Do what you could do. get the marks. So, most likely it's four or five marks for this. Even if you hit a sticking point halfway through, that's half the marks. That's better than none, right? Okay, let's take a look at what they given us here. So, tan theta by sin theta over 1 minus cos theta and that's identical to 1 + 1 / cos theta. That's interesting.
All right.
Um, a proof like this can come. This is a mass paper. This is 2014. So yeah, it could it can and has come before, but the ones more recently have been a lot easier. But you never know what they're going to bring or how they're going to bring it. So you don't want to underprepare.
Cool. Okay. So to that end, let's start by say okay. Okay. So, we're saying okay. So, the proof.
All right. We're going to take the left hand side.
Sorry. One sec. Uhhuh.
Did the community get back to you?
What? Sorry.
Uhhuh. Rad. Hey, what going on? Sorry.
So far, boy.
All right. Cool. Okay. So in that in that fraction so sorry let me write it.
So tan theta by sin theta all over 1 - cos theta right so there's nothing squared here so we carry attack from that one so the only one we can kind of attack is the tan right okay cool so let's let's go from that side so with that so we could have >> correct all right Daniel how you with this on. All right. All right. So, um what's going to happen here? Right.
So, check this. Right. So, this is another case where you could if you need to um what's the word I use decompose the fraction.
Right. So, tan theta / 1 by sin theta / 1 by 1 / 1 - cos theta. And of course tan theta becomes what?
Sin theta over cos theta.
And multiplying that by sin theta right.
Okay. So when you reconstitute the fraction um sorry yeah and you multiply numerators together. So sine by s will give us and of course well multiplying by one changes nothing right now in the denominator you have cos by 1 minus cos so we're just going to put that like a coefficient in our bracket right but now in the numerator there's the presence of a square sin squared so guess what we could do we could attack that so what what will that turn into aha and 1 - cos^ squ is a difference of two squares.
So guess the next move correct factoriize it.
Yeah. There you go. There you go. the drill.
Right? So we have a point we reach a point we have 1 + cos theta by 1 - cos theta all over cos theta by 1 - cos theta. So guess what's going to happen here now?
Numerator and denominator will cancel leaving us with uhhuh and here we decompose the fraction.
How do we do that when there's not multiplication? Well, it's addition, right? So if you think about it, before we add, we'll have two separate fractions. And to add them, we have to have a common denominator. So all you'll do is you'll take each aspect of the numerator and express it separately over common denominator right and guess what cos theta over cos theta will give us one right we have 1 / cos theta + 1 which of course is the same as 1 plus 1 over cos theta so if you want you can shift it around right yeah g very good yeah so we're good there right so yeah so you You got to look for the look for the vulnerabilities. Look look for where you can apply the identities and like I said there's only two we have tan theta is sign over cos and sin square plus cos square is what and any rearrangements thereof. All right. Okay. Right. This one is a little involved because of the presence of this item here that was apparently never on the CC syllabus.
That's the inverted that's just one over course. So what are going to do? We are going to rewrite that for you. Right.
Um, all right. So, it's it's actually supposed to be one over, right? So, hold on. Did I say there was a twist here?
No, sorry. It wasn't this question. It was this question as a twist. A special technique, right? So, SE is actually the inverse of cos, right? Plus tan theta, right? And this is identical to 1 minus sin theta all over cos theta. Okay. So what what I could suggest you do here is again you could take the left hand side right and you could work with with the denominator and of course the vulnerability there. The point where you could apply something is the tant right so start there. Jaden. No, Jaden.
Apparent I thought it was when I came into this back in when I started teaching CERT mass back in like well for the 2012 year. I thought it was I used to teach it but apparently it's not right.
So in the first ever CC admass exam they brought something that wasn't the although they expect you to know. No they didn't.
The 2012 paper was the first of all mats here. I I have some mark 2011 that was actually the specimen paper.
>> It said yeah I agree it's still good. So send back us. Yeah. Yeah. Yeah. Yeah. We prepping for we doing some trick proofs here.
So, right. So, we're working with just the denominator for now. So you have 1 / the sum of 1 / cos theta and sine over cos theta. Right? So because they have the same denominator, all that means is we're going to just kind of add the numerators together over the common denominator. Right?
All right. Are we okay up to that point?
>> Okay. Now the interesting thing here is you have one over a fraction. We saw that a bit earlier. I think in the first question 1 / tan ended up being 1 / s cos. And when you have that, what happens? you invert the denominator. So basically what's going to happen here is we're just going to get cos theta over 1 + sin theta. And this is the sticking point here, right? Jaden, you know the thing, man. Very good. Very good. Very good.
All right. Now, I'm going to show you the trick here. Right. You're familiar with with with SS, right? You know in search sometimes you have to do something called rationalizing a denominator. So let's say you have like a roo<unk>3 minus 2 or something. So what you do is you end up multiplying by what we call the conjugate. You multiply by whatever we need to change that denominator to a difference of two squares. So if it was let's say was roo<unk>3us 2 the conjugate be roo<unk>3 + 2. So you multiply by roo<unk>3 + 2 over roo<unk>3 +2. So what what's going to happen here is you're going to multiply by 1 - sin theta over 1 - sin theta. So this would be that would have been the sticking point there. Right? That first that first exam.
>> Yeah.
>> Right. Multiply by the conjugate. Oh. So J basically what it does Jaden is it facilitates um the formation of a difference of two squares in your denominator which will allow you to further manipulate uh and well maybe simplify may be the better word. Right?
Right? So it's just basically finding a way to continue what he was doing. Well, to continue the working, right, in a rational, logical way. It might seem arbitrary like where that come from, but um I'm not even sure where it come from, but it is a technique, right? That's just been kind of passed down, right? We check it. Uh if you kind of put the fraction together, right, you multiply your numerator. So you have cos theta by 1 - sin theta.
And then you have well of course because we did difference of two squares here sorry we did the conjugate we know that's going to be difference of two squares so we could just put that as 1 minus sin^ squ I mean if you want to go the route of put in 1 + s by 1 - s you could do that and then show that that becomes 1 - sin^ square and of course as we know 1 - sin^ square becomes what mhm co right and guess what happens now the cause in the numerator ator could cancel with the cos squ in the denominator leaving us with what >> minus sign all over co and is that what they wanted us to show?
Yep.
All right. Okay. Right. There's one more which was actually from the specimen paper. So if you're feeling fancy, you can try it. Right. Um at least at least there's no inverted trig ratio here. That that at least that I would give them, right? Okay.
All right. So check this, right?
Um it should I should erase this piece here. Anyhow, so small thing. Um it's just that the presence of the minus in the front here could be a little it Some people didn't even see that minus there initially. I didn't either, right?
Anyhow, but this actually to me is lifted from the um the yellow and yellow and green text the uh Jaden. Oh, thanks J. Well, I doing it a long time now. So, by now I continue to weld something wrong. But over the years, I kind of refine my approach. It used to be different. Used to be a a little more complicated, right? But over the years, you had to learn from how when you do things. All right. So, a lot of people's first instinct is to apply the square. I would say hold off on that. simplify the inside first. Right? So of course what that means is we're going to replace tan theta with what?
>> Right? So start there.
Actually, maybe I could maybe I could go Oops, that's maybe I'll go horizontally this time.
All right. So, now that we have sin theta over cos theta minus 1 / cos theta, we could simplify inside the bracket, right? I think we did that just now too, right? because they had a common denominator denominator. So let's go there. So long story short, all that will happen is you'll have sin theta minus one all over cos theta.
All right. And now when you apply the square, when you square something, what are you actually doing? Multiplying it by itself. When you multiply a fraction, you multiply numerators together and then denominators together. Right? So long story short, what's going to happen here is you're going to end up with sin theta - 1^ 2 all over cos square theta. Now if you look at where we go in, what's the only trig function inside of there? Sin theta, right? So how do we quote unquote get rid of the co square?
Uhhuh. Right. What we going to convert it to?
>> Uh-huh.
And as you know, 1 - sin square is the difference of two squares, right? So here what we're going to do, we're going to we're going to express the numerator as sin theta minus 1 by itself.
And the denominator we're going to express as the factorized difference of two squares.
Very good. Yes, that's correct. That's correct. 1 - sin square. Yes. So, of course, a a square minus b square becomes um a + b by a minus b. So, or a minus b by a + b.
All right. Now this is the sticking point here because if you look at where we going they have sin theta minus one and then over sin theta + one. Um that is J.
>> Hey J. How you doing?
>> All right. Cool. Yeah.
>> So um yeah. So we doing a little proofs.
I did proofs earlier today with um Luna.
But what happened was um our boy Machine was giving trouble. So I started the first session about 40 50 minutes late.
So I only got through like this first set here. Then I had to start back a second session and do some of the compound angles and that right. Um and then the two of the other girls coming back later to do to do um stats again cuz they they miss it yesterday. So here we stick around for the rest of the class, right? I'll send in the replay and then um you could watch that during the 7 to half eight time slot, right? Um cuz um the girls want to work the stats.
So I'm working with them.
Yeah.
>> All right. Cool.
Yeah. So, um, so we started off with the, um, some basic notes on on the proof that we really only have two identities we use in CC maths. And then, of course, you have different slightly different versions of them, right? Uh, not not counting the double and compound angle formula, that's a whole different batch, right? So, we started from 2024, we were backwards. Um, we saw that 2023 and 2018 were the same question. So we didn't work it we didn't work it twice right and if you notice I mean as we as we went backwards in years that's when the workings start to get a little more right the funny part was the the more recent years have been quite relatively easy um in a manner of speaking of course right anyhow right okay right so you get to this point here um we have a little quote unquote problem because we want to end up with something like this right so I mean we have the sin theta minus one And the 1 plus sin theta you could write as sin theta + 1. But somehow we had to get rid of these things here. But they have this minus sign in the front here. So what happened there? So this is this is the trick, right? In order to be able to cancel the those two, you have to factoriize either one of them using minus one as the common factor. So for example, right? If you have sin theta minus1 and you pull out minus one as a common factor, what do we do? We divide each term by the common factor, right? So the positive sin theta when divided by negative 1 will give us minus sin theta and the negative 1 / will give us plus.
Right?
So you could and and on top of that what would happen is you could always switch around to get one minus sin theta.
So what's going to happen right is um so when we come down here now if you pull out the minus one which could just be a minus sign you get 1 minus sin theta right by the same sin theta minus one all over 1 - sin theta by 1 + sin theta and guess what happens Yep. The 1 minus sin theta cancels numerator and denominator leaving us with just the minus sin theta -1 over the sin theta + 1 right because 1 + sin theta and sin theta + 1 are the same because addition is commutative.
I'm not changing the minus sin theta 1us theta.
>> You you could I I just wrote this there as an alternative, right? But they wanted us to show it this way. So it's it's better to show it the way you they wanted. I was just showing that it could be written that way too, right? So if you got at that point, you'd have to do all the extra step to facilitate.
>> Yeah.
>> Yeah.
Yeah. I I I um I've experimented with it. I've gotten it in different ways. So yeah, all that valid. All right. Right.
So interestingly enough, those are all of the trick proofs without going into double and component. So um I don't know how you're feeling with it now. I know you had some reservations before we started.
>> A little better.
>> Yeah.
>> All right. Good.
>> The older ones are >> the older ones. Yeah. But as you can see, they're not bringing them soon. But there's no guarantee about how they're going to bring what they're going to bring. They might decide. Let me go back to some nice old Let me looking at Cape Dex.
Oh boy. Anyhow. All right. So, now we're going to discuss the compound and double angle format. Right. So, I'll print this off.
>> Right.
>> Yeah.
>> You have it.
Yeah.
>> All right. Well, maybe I mean maybe it's one of these questions. Yeah.
>> Okay. So, almost like this then.
Yeah. Yeah. No problem.
>> I hear you. I hear you. Those those need work.
>> Yeah. Pythagoras is is a a valid way to work it because I mean it's right angle triangle. So once you know how to apply it it's um it's it's a valid way. All right. Okay. So you all don't have to know the derivation of these um expansions. Right. I I was curious. I used to sit at night trying to figure it out and then I read a book and then use the unit circle and rotating the arm and similar triangles. I was like way this is fun. Anyhow, um I think if you go to to to cape and do I think pure max, I think it might have it might have to I'm not sure. Um but there are different ways to do it. Anyhow, um basically all of these items here are going to be on your formula sheet. So you don't need to memorize them. I would always say know all of your formula.
Don't depend on the formula sheet. We spotted an error in one. I'm like, how is an error in the formula sheet? Don't you just copy and paste it every year?
>> So just one year.
>> Yeah. Yeah. I think it was like 20 either 2013 or 2015. I think it was 2015. Anyhow, um right. So basically what we're going to do is we are going to look at the question. We're going to see okay what is it looking for? Is does it does it does it give us a double angle or a compound angle right? The double angle simply means instead of using a plus b where a and b are different it did a plus a right which is 2 a and then you had a a here and a a here and that sort of stuff right?
Obviously you can't do a a minus a because then that's that's just zero. So that wouldn't go anywhere really, right?
Okay. So this is Oh, wait. I need to give you a little booklet to do this, right? Give me one sec.
>> It was doable. I don't like that they didn't bring any circle geometry. I don't like that functions every single year.
>> Yeah.
I looked through it. So I just started by looking through it and I was like >> I was confused. I wasn't anything.
>> I didn't like um that yeah they had no circle geometry and that they kind of overindex on matrices of transformation because they brought it in in place of circle geometry and in the matrices question they brought it again. I'm like you all have so much of us that I must attest.
>> Exactly. Like >> keep it in one place.
>> I know.
>> So I was we spend the time with each other.
>> Sure. At least we that that much we can say kind of work in our right.
>> Yeah.
>> Right. And here the next thing right the next thing was that question.
Why is there an SDA question on a CC paper? What what they really doing?
>> Yeah, I don't understand. And then they had so much so many other things they bring forward and they had al already brought out ratio questions early on in the consumer section.
>> Yeah, exactly. Two ratios questions and they brought into the patterns as well.
They brought >> they brought they brought two sets of patterns for you all.
>> All right. Um I do mine but at the same time >> Yeah. with the diagram. I mean people I don't like the diagram to draw because I can't draw right fine but still some people like to draw and if you get some marks for that >> everybody is true that the pattern they come every year and they just didn't bring it to >> so they've broken the pattern.
>> Okay. So this is from 2022. So it says hold on one second. Uhhuh.
Sorry. Give me one second.
What is this now?
Okay, cool. Cool. Right. Okay. So, here it says show that cos 2 theta is equal to this. Now, the funny part about it is that actually is on the kind of formula sheet here, right? But they want us to be explicit about how we got there.
Okay, cool. So, here what we're going to do, right? Okay. So cos 2 theta is equal to cos of theta + theta which basically is the cause of the like like you can mirror with the a + b right and if you go back to your formula right the first item in the green here cos a plus b is cos a cos b minus sin a sin b.
So all we're going to do is we're going to we could write it below. So now you don't have to write the one in the green. You could go straight to to the pink. I'm just writing it so we have a little template to follow from below.
Right. Right. So now in the pink which is the actual thing they want us to do.
So we'll have cos theta by cos theta minus sin theta by sin theta. Right? Let me skip a little line. Right? So what's cos theta by cos theta?
cos² theta. What's sin theta by sin theta?
sin^ 2 theta. Okay. And now look where they want us to go. They want us to only have cos squ.
So it means we have to somehow kind of get rid of the sin squ. And again that goes back to our primary identity, right? Sin^ square + cos square is 1. So sin^ square is equal to 1 - cos^ 2. So you're going to swap out sin^ squ and put in brackets 1 - cos^ 2. And of course when you drop the brackets what's going to happen there plus cos square theta right if you're feeling like you want to switch around some stuff to put the cos squares together and of course cos square + cos square is there you go right and then we done right so relatively straightforward and simple how we feeling with that one >> all right cool J you feeling with that one that one's all Yeah.
I didn't hear you. I saw the thing like that. But your mic check and see if it working the plug.
>> Yeah.
>> Yeah. Yeah. I hear now.
Right. Okay.
>> All right. Cool. Right. 2015. So 2015.
Hold on.
>> Yeah. Okay. So 2015 we have two double angle formula. We have a sin 2x in the numerator and a no sorry just one. Okay.
But something interesting I want to point out here. Right on the left hand side, what's the denominator?
Oh, sorry. It wasn't that.
>> Oh. Oh, wow. What I was doing?
>> No, sorry. We're doing 2015. Sorry.
Sorry. Sorry. Yeah. So, yes. So, you have a a double angle in the numerator and your denominator. And look at where they wanted to go. Tan theta. And that's like I said, that's one of our two primary identities. Stan theta is sin theta over cos theta. So somehow somehow we had to kind of cancel stuff to almost get rid of everything in the yellow. So it's just sign over co right? But to cancel down we had to expand right. So if you go back here sin 2 theta is 2 sin a well 2 sin theta cos theta. So let me um let me expand.
Right.
Okay. So proof, right?
So we're jumping in one time. So we're going to replace sin 2 theta with two sin theta cos theta all over right.
Okay.
Are you sure?
>> Right. And in the denominator now you have 1 + cos theta and then you have cos 2 theta.
All right. So something telling me to stick with just cos in in the denominator. So we just what we just saw in the previous questions 2 theta is 2 cos square minus one. Let me use that.
So plus 2 cos square theta minus one.
All right. So, what we're going to do, we're going to fiddle around with that denominator a little bit, and we're going to um we're just going to cancel the one and the minus one, right?
So we're going to end up with just sin theta + 2 sin theta cos theta all over cos theta + 2 cos^ 2. And if you're feeling fancy you could you could even write cos square as cos theta by co. There's no need to do it. I just wanted I just wanted to show you something a kind of similarity between the numerator and denominator right? If you notice, you have like a sign here and then then you have a two sine something, right? You have a cause here and then you have a two cos by cause, right? So the next step here is to factoriize. In the numerator, your common factor is s.
And when you pull out sign, guess what?
You get 1 + 2 cos theta.
And in your denominator, your common factor is cos.
>> And when you pull that out, you're going to get 1 + 2 cos theta. And guess what happens there after? There you go. That cancels with that. And you're left with sine over cos. And sine over cos is tan theta.
All right.
G. Now we going with that.
All right, Elliot, how we feeling about that one?
>> Yeah. All right. Right. So, back to 2015. So, what I was showing you here is that you have 1 + sin 2x / sin x + cos x and it's equal to sin x cos x. So to me, the only way that could work is if the numerator is actually equal to the square of sin x plus z. Right? Now, this is where I I almost took a a serious route here where I went all over the place. Right? Um but hold on.
>> Okay.
So the sin 2x we are going to expand into its well alternative presentation what J what the previous one it I mean okay yeah I understand Jen I doing this since like 1996 so like we're talking about >> no sorry no is a different Jaden >> sorry >> oh you did Yeah.
>> Oh, okay.
>> What was the question?
>> All right. I know have different forms, right?
>> Right. That's correct.
>> Um like multiple times in exams. I would go wrong for first I try >> right. So yeah, a a trial and error approach in an exam if it has time could work time and space, right? But okay so let a little tip I could give you is and again sometimes this is just kind of after a while in intuition because I've worked so many of them right but you see how I have tan theta here right we know tan theta is s over cos now in the numerator you have a a sin 2 theta which you end up with a cause in the numerator but basically the majority of what is here is sign right and in the denominator when I expand this I say okay maybe I want to try and only cause in my denominator. Right? So, it it has to do with with looking at what is given to you and looking at where you had a go and trying to find some kind of connection. Sometimes it might be a little easier and sometimes you might see nothing and they just had to go one step at a time in the dark and hope you don't bounce into a wall or fall off a cliff. But that that is the risk here.
But again, if you work enough for them, you'll have a better intuition. But I mean, the exam is Monday. So I know how many you I actually have I have a a video up on YouTube from I think is it 2019 where I have like 60 something and I work from different textbooks and keep books and there some of them I took real long to get out. Um I don't even know how I all that right now but but yeah there if you ever looking for some extra ones to work you could always check that video. If you can't find it let me know.
I'll send it to you.
>> Yeah no problems. All right. So here, so we know that sine um sorry sin 2x when you expand you're going to get 2 sin a cos a right and the trick here is to expand one.
One is equal to what?
sin^ 2 + cos².
And of course sin 2x is equal to 2 sin x cos x.
This is all over sin x + cos x.
Now what I'm going to do is a very very slight rearrangement right and it's based on it's based on the perfect square. Right?
So if you have a + b all squared and you expand, you're going to get a 2 + 2 a + b ^ 2. So if you consider that like a is basically s and b is basically cos, we could kind of see what going to happen here, right?
>> You're going to get sin squar plus correct. Yeah. Yeah. No, that's correct.
What did I say at the outset? I said for this to work that numerator has to be the square of the denominator.
Difference of squares. Oh, put your perfect square here. Small thing. Right.
Right. And again here in the numerator sin^ 2 x + 2 sin x cos x + cos 2 x can be factorized as sin x + cos x all squared all over sin x + cos x. And guess what?
Correct. sin x cos x sin x + cos x will cancel with the with its square giving us just sin x + cos x which is exactly what they wanted us to show Yeah.
>> Paper one they do come. Yeah.
>> Okay. So like if they don't bring it in paper >> boy the paper one kind of it just repeats. Yeah. So I mean by the time we we get there and we do a few papers you'll kind of know it inside out. So um don't don't worry about it too much.
>> All right.
>> Yeah. All right. Cool. There you go again. 26 prove. Okay. So here we do have a double angle one. We have a compound angle. So the theta plus alpha is like the a + b.
Right? Now they want us to show this is equal to tan theta plus tan alpha. Which is interesting because we know tan is s over cos. Right? So somehow we had to get sin theta over cos theta and sin alpha over cos alpha. So if it was just plain old distributive law we could have well no we couldn't do that because then it would just be multiply. Anyhow, here what I wanted to do. I wanted to expand this using the identity the expansion and then you'll be able to decompose the fraction.
Easy one. All right, let me see.
We have sin a cos b a sin b.
All right. To the end of sin theta cos alpha plus cos theta sin alpha all over cos theta. Right. So now we break up the fraction. So we put each aspect of the numerator separately over its common denominator.
And guess what happens here? We could cancel off some stuff. Cos alpha with cos alpha cos. And so we left with what?
Sin theta / cos theta plus sin alpha over cos alpha.
which is equal to theta plus alpha.
So how we show what what they require?
>> Yeah. All right. Cool. All right. J, are you okay with that one?
>> Yeah.
>> All right. Lovely.
Okay. All right. So, we have something slightly different here. It says a right angle triangle XYZ has an angle theta where sin theta is roo<unk> 5 over 5.
Without evaluating theta, calculate the exact value in third form if applicable of cos theta and then sin 2 theta. Now we know that sin 2 theta is equal to 2 by sin theta by cos theta. So that's probably why in part one they want us to work out cos theta right now cos theta we have sin theta we want to find cos theta so what we have to do here is we have to use the primary either the sin square plus cos^² equal to 1 and reverse it. So basically what I saying is if you want to find cos right you could find the square root of that.
So cos^² is 1 - sin^ square. So cos is the root of 1 - sin^ square.
>> There's the one here. Ah 1us sin square theta.
All right. So let me give you a little a little chance to work out something there. Now it says it's a right angle triangle. So we know that the angle will be um acute.
It's a positive value.
So, take a little swing Oh, that's what's going on.
Sorry, there's something in someone this Yes.
Uhhuh.
>> When she went and picked it uph >> right >> but like I didn't want to take a shot.
So she >> right >> and then she replaced it on the sometimes like it just >> and now it's just not >> as a new calculator.
>> Okay. So, when they say replace >> the battery.
>> Oh, the battery. Okay. Well, that sounds like it fell and the screen got damaged or something inside got damaged.
>> So, the battery probably not going to help unless I mean did you try resetting it and see if that helped? Sometimes that could that could right. So, we're going to get five over So, so 1 is 25 over 25 by the way, right? So, that's going to give us 20 over 25.
So, that's<unk> 20 over<unk> 25.
Oops, sorry, my bad.
Right, which gives us um So that's 2<unk> 5 over 5.
Right? And now here we're going to have 2 by<unk> 5 over 5, which is same as that value 2 by<unk> 5 over 5. Of course, 2x2 will give us 4. R 5 by<unk> 5 is just 5. 5x 5 is 25. So we're going to end up with again 20 over 25, which is just 4 fths.
I have it as four.
>> Okay. No, that's fine. So, right four over five, right? So, what would happen is you will get 2 over<unk> 5, right? Which you could use and then that is correct. And then what it would so so my version of 2<unk> 5 over 5 is the same as that. It's just that all I did if you consider it 2 over<unk> 5 then multiplied by<unk> 5 over itself >> will give us 2<unk> 5 over 5 which yeah so it it it should not be a problem.
All right J with that one.
>> All right cool. All right, next one. You have any questions?
Right.
Go again. Right. So in this one now given that cos m is equal to 24 over 25 and that angle m is acute which means it is less than 90 between 0 and 90.
Determine the value for tan 2m.
Let me start with what tan 2m is. Right.
So tan 2 m is a double angle formula right tan 2 m is 2 tan m over 1 - tan^ 2 m.
But the thing is to get tan m we have to have what? Sin m over cosm.
Now we have cos m now. So that so basically what we're doing here is like the reverse. In the previous question they gave us sign and we had to find cos. Now they gave us cos and we had to find s.
All right. So cos m is equal to that. So therefore sin m is equal to 1 sorry the square root of 1 minus cos^² m. All right.
Give you a little time to try and work it out there and then we'll go from there.
So the 1 - cos^ squ might be a little involved.
That's 1 - 24 over 25 then squared. So that's 576 over 625. Oops.
Where you going? Where you going?
And that will give us 49. The root of 49 / 625, which gives us 7 over 25.
All right. Now, you could use that up here. And when you get that, you can use that up here.
Very You cross.
Okay. Right. So sin m over cosm is the 7 / 25 / the 24 over 25. And of course when you're dividing fractions was the protocol you change divider multiply and you invert the divisor right the 25 will cancel and then you have 7 over 24.
I use that up in the final stage here.
That's two by um that's 7 / 12 then divided by 24 squar is 576 six.
So I have 46 really.
So that's 5 20. No, five 27, right?
and 12 into that will go 48 * 48 by 7 50 by 7 is 350 - 14 is 336 yeah I don't know if that could cancel all right so that there so we there all right Okay. Yeah. Jaden, talk to me. You good there? Yeah. All right. Lovely.
Okay. So, now we get to the kind of interesting part. I mean, it's all interesting, but the the the part where we had to use some SS and radians and that kind of thing. So, I have I have a nice little cheat sheet for your head, right?
Um, so I'll print it off and I'll I'll kind of do it over from scratch so you can see kind of see it in motion.
I'm 34.
What is this? One second.
Huh?
Right. Okay. So, we are required to know certain exact values for s cosine and tangent for certain angles. So these are your starting points here, right? So like I was telling Luna earlier, basically you could think of it like um 0 0 30 69 like it goes up in 30s, right?
And 45 is kind of along for the ride as it's exactly halfway between 0 and 90.
Now there's a pattern for these that I saw on Tik Tok a few years ago that I like and to me it it makes it a bit easier to remember. Right? Each of these values is is all going to be a root of something all over two, right? They're all going to be root of something all over two.
And those things are actually pretty easy to remember because they're just five consecutive numbers starting from zero. So 0 1 2 3 4.
And the good news about it is um the cosine values are the exact same ones but in the reverse order. So this one starts with the roo<unk> of 4 /2. This one is the root of 1 / two roo<unk> of 2 over2. And we'll notice that s and cos 45 are ex the same or equal. That's correct.
And this will be the root of 0 / two.
Right?
Now let's see if you could simplify some of those things. So of course root of 0 is 0 and 0 / anything is 0. The root of 1 is 1 and 1 / 2 is just a half. Root of 2 over2 could stay. I used to as 1 / roo<unk>2 which is just it's a kind of different version. Roo<unk>3 / two could stay roo<unk>4. What's the root of four?
That's two. And 2 over2 is 1. Right? So down in the green that's equal to one.
Sorry, it's supposed to be three, right?
So that one doesn't change that one. So this one is a half and this one is zero.
And to find tan, tan is equal to s / cosine, right? Cool. So check it. Sine of 0, which is 0 / cosine of 0, which is 1. 0id anything is 0. Here you have a half / roo<unk>3. So when you divide in fractions, the protocol, change, divide, multiply, and invert the divisor. So you'll get 2 over roo<unk>3. And the two will cancel with the half here leaving us with 1 / <unk>3.
In this one, <unk>2 over 2 / by itself will give us 1. 10 45 is one. Here, oops, sorry, my bad.
Yeah. Here for the 60° one. Again dividing this by that. So you'll invert the half. The two will cancel leaving us with just roo<unk>3. And here now 1 / 0 anywhere by 0 is undefined.
And if it if you check this, you know how a quadratic function is symmetrical about its maximum minimum point and the values kind of repeat.
>> Yeah.
>> Here it they don't repeat. They invert.
Check it. Right? If you invert zero, put one over zero, what you get undefined.
If you put one over this, you get that.
And then one over one is itself. Right?
And here. Now the other thing we have to put in here is the radian versions right now the rad all you have to do is divide the angle by 180 or express it as a fraction over 180 and then multiply by pi 0 over anything is zero and 0 by anything is just going to be zero right um let's start with 90 sorry 90 over 180 is what fraction as a half and a half by pi will just give us p<unk> / 2 60 over 180 6 into 18 is 3. So 1/3, right? This is p<unk> / 3. 45 is actually half a 90. So half of a half is a quarter. So this is p<unk> / 4. And then 30 is a half of 60.
And a half will give us a half of a third will give us pi over 6. Right?
Now check this. Right? If you notice the numbers going backwards here, they're almost consecutive. 2 3 4 and then all of a sudden six. All right.
Okay. So with that now we have to be able to do certain questions with the compound angle formula with the split up angles and what is it in third form. All right. So we're going to start with one where the angles are in degrees and then we're going to try one I'm going to do one with you where the angles are in radians. Right. And then if we have time cuz I watching 6:42. I was supposed to be on break from half 6 or 7 but that's all right. We have things to do. Right.
>> Yes.
>> Yeah.
Right.
So given that cos 30 is equal to that and sin 45 is equal to that without the use of a calculator evaluate cost of 105. Okay.
What I want you to be able to do here is to use your compound angle formula. So we do have and correct right now let's just for a second consider 30° and 45° right by themselves.
30 and 45 will give us 75. So let's just say instead of cos 105 they wanted cos 75. Right? What we could do is we could express that using the compound angle formula.
And then when you use your expansion you get cos 30 by cos 45us sin 30 by sin 45.
And then you plug in your values and simplify. Right? Let's check it out. So if if you did cos of 30, that's roo<unk>3 over 2. Cos 45 is <unk>2 / 2.
Sine of 30 is a half. Sine of 45 is the same as cos 45 roo<unk>2 /2. Right? And when you multiply and simplify roo<unk>3 by roo<unk>2 is just roo<unk> 6. 2x2 is 4. 1x <unk>2 is <unk>2. 2x2 is 4. And because the fractions have the same denominator, we could combine into a single fraction.
Right? So that's kind of what they wanted to do. Right? Now the weird thing and then I had a problem with this question. I'll tell you why. The space they gave you was insufficient in my opinion to the question. Right? Unless they wanted you to actually not use what they gave you. Let me explain. Right? So let me just kind of box off this side here as the example. Right?
105.
Now 30 + 75 30 + 45 is 75. Right? Hm.
How do we get to 105 then? We can use 60, right? So, what we're going to practice here is using 60 as opposed to 45, right? So, let's do that. So, cos 105 is equal. Now, you can put the 60 first or the 45 first is addition. It's commutative. It don't matter, right?
Let's put cos 45 + 60°.
Right? Now, let's use the expansion. So cos of a plus b is cos a cos b minus sin a sin b right.
Oh that child again boy.
Okay so far so good. Right now cos 45 is <unk>2 / 2. cos 60 is actually a half.
Sin 45 is the same roo<unk>2 over2 and sin 60 is actually <unk>3 /2.
So in the<unk>2 by 1 is just <unk>2 and of course 2x2 is 4. Um <unk>2x roo<unk>3 is roo<unk> 6 over 4. Oh so interestingly enough this is kind of like the like the reverse of that the negative of it actually.
Right.
So, if you want, you can drag that down.
Oh.
Oh, I didn't print it off here. No.
>> Oh, I know. Oh, well, I guess that would because now you actually have the the example, right? Okay. So, here what I'm going to do. Let me do the next example one time. Um, now I want you to kind of follow along with that one. Okay. Good. Okay. All right. All right. All right. All right.
Um, think t All right.
So, we're going again. All right. So, here given that sign of pi / 6, so pi / 6, right? P<unk> / 6 was 30°, right? And then cos of 30 is <unk>3 /2.
So they want cos of x +<unk> / 6 and they want us to show that is equal to this where x is acute. Acute just simply means the values of s cosine and tangent are all positive cuz acute angles are in the first quadrant 0 to 90 and in the first quadrant all of the ratios are positive there. Right? So that's a it's not just that's not a random thing they put. Okay cool. So let's expand this.
Right? So this is a cos a plus b. The expansion of cos a plus b is cos a cos b minus sin a sin b. All right, cool. So cos x cos of p<unk> / 6 minus sin x by sin of p<unk> / 6.
Right? They gave us that cos of p<unk> / 6 is <unk>3 /2. So we have cos x by <unk>3 / 2 minus.
So sine of p<unk> / 6 is a half, right? And what they did is they use a half as the common factor.
So when they pull out a half and then divide, right, they're just going to get So if you divide this by a half, long story short, it basically pulls out your denominator, right? And cos x by roo<unk>3 is the same as roo<unk>3 by cos x. And of course, when you divide by a half, you're left with sinx, right? So a nice easy one to kind of start with.
>> Yeah.
>> All right. Cool. Next, 2017. So 2017 is a little more involved, right? Right?
You might have to go on the next side of the page to do the work. Right? So they give you a whole set of thing here.
Right? They give you, right? So sine of p. So sine of 60 is that cos of 60 is a half and then sine of 45 is equal to cos of 45 which is <unk>2 /2. So they wanted to show that cos of p<unk> / 4 -<unk> / 3 all over s of 2 pi. So 2 pi over 3 is actually p<unk> over 3 plus itself. So this this down here comes like like a 2 a where a is pi / 3.
Yeah.
>> Okay.
>> So they wanted to show here that that's equal to this. So basically what you're going to do so this is like a cos a minus b and this is a s of 2 a.
So what you could do use your expansions after you expand plug in your values and simplify to get hopefully So, if it's still working, that's fine.
Just allow me to start.
All right.
Cos of pi / 4 is <unk>2.
Cos of pi / 3 is a half.
Sine of pi / 4 is <unk>2 / 2. Sine of p<unk> / 3 is <unk>3 / 2 all over 2 by. So sine of p<unk> / 3 again is roo<unk>3 / 2 cos of p<unk> / 3 is a half.
Okay. How we feeling with that so far?
>> Yeah. All right. So we're going to end up with <unk>2 over 4 in the numerator and then <unk>6 over 4 as well. And then down here what's going to happen is actually you could cancel the two with the half and just get roo<unk>3 over two. And then what's going to happen here? You're going to end up with the numerator has fractions with the same denominator, right?
And of course, when you're dividing by a fraction, you change divide to multiply and you invert the divisor.
And two will cancel with four two times, right?
So of course, now your numerator, it doesn't change. You still have the same roo<unk>2 minus<unk> 6. And now two by roo<unk>3.
>> Oh no, you Yes. Sorry, you are correct.
Yes. Yes. Yes. Wow. Thank you.
Wow.
And look at that. Did we get what they asked us to get? Yeah, we did.
All right, J.
Yeah. All right, cool. All right, now we have about five minutes left.
>> Yeah, let me just go have a boy. Why are you feeling?
>> Um Oh, what you doing? You doing the stats too or >> Yeah.
>> All right. Who is this? It was Jaden and Zabi did the start with me yesterday.
Okay, cool.
>> Yeah.
>> Yeah. Yeah.
>> Yeah. Well, I guess I had to repeat that. So you um so what what to send for you Jin what what what it is you need to he made a sent for you so you could do it or um the earlier part of the same replay. Yeah. Yeah. Okay. Okay. Yeah.
That's a good Yeah.
>> Where did you guys >> Oh, this this morning I I did the same thing um with Luna. But what happened was um um the first session I started late and then the computer was giving trouble. So, I only got like 40 minutes in that session and all all I did was at the first piece here and then when I did the rest of it, I didn't even get as far as I got with you guys, right? We didn't even work any questions here, right? Um, so you guys did did more than she did.
So, but you you missed the first part of this, right? So, what I'll do is I'll I'll send the replay for this session.
So, watch like the first the first set of proofs and then you should be okay.
>> I see. No problem.
>> Yeah. All right. I think we have time for one more, right? Um so given that sine of p<unk> / 3 is <unk>3 /2 cos of p<unk> / 3 is a half and then s of p<unk> / 4 is equal to so basically they gave us the same information as across here sine of pi over 3 cos of pi over 3 and then s and cos of p<unk> / 4. All right.
>> Yeah.
Yeah. So evaluate without using calculators the exact value of cos of 7 pi / 12. So you're watching what they give you. They give you p<unk> over 3 and they give you p<unk> over 4, but they want 7 pi over 12, right? And to get that, you had to know that, okay, if I took p<unk> over 3 and I added p<unk> / 4, right? So 3x 4 is my common denominator, which is 12. And then you multiply each numerator by the opposite denom. So you're going to have 4 pi + 3 pi. That's going to give us 7<unk> 12.
So basically if they wanted to find cos of 7<unk>i / 12 this is cos of<unk> / 3 +<unk> / 4 and that's a cause of a okay so expand it plug in the values and let's see if we could wrap it up with There's a cause of a plus b. So we'll have a minus sign in the middle, right?
So cos of a by cos of b minus sin of a by sin of b.
Okay.
So cos of 60 is a half cos of that will be <unk>2 / 2 minus sin that is <unk>3 / 2 by again <unk>2 / 2 that's going to give us <unk>2 / 4 -<unk> 6 / 4.
That's just going to give us <unk>2 -<unk> 6 all over 4. Yeah. All right.
The last question was from the specimen paper. Um, we don't had to work that one. Um, hold on. I know what to do now.
I should have cropped this one out.
Yes. No. Crop some more.
All right. Okay. So, yeah, just hand this one real quick and then we'll call that George.
>> Um, that would be all of the questions.
You would have done all in a twoour space, which is pretty good.
>> I have one question.
>> Yeah, go through. Why do they specify?
>> So when x is acute, so what that means x x is acute means between 0 and 90°, right? If you remember the quadrants, right? From 0 to 90, all of the ratios are sorry. Yeah. Tan sorry sine, cos and tan are positive. Right? In the second quadrant, what happens is only s is positive. That means cos and tan are negative.
In the third quadrant, s and cosine are negative. Only tan is positive. And in the fourth quadrant, uh sine and tan are negative and only cos is positive.
So when they say that right, they mean that when you find when you like use the exact value, you don't have to worry about whether it's going to be positive or negative. It's definitely going to be positive. It's just to give it like a safety net.
>> Yeah.
Yeah. So <unk>2 over two turns out to be the common factor.
So you factoriize that and you end up with sin x - cos x in the bracket.
Yeah.
Yeah, >> that's okay. We We know what it is.
Okay.
All right. So, we're going to take a short break there. All right. Yeah. So, J what I'll do is let me uh let me log off here. I'll send you the replay and you can watch back the first the first section of the um of this one to get the the the more run-of-the-mill proofs, I guess you could say. All right. Um because they were here for the the more technically technical involved stuff.
All right.
>> All right. Okay. And um tomorrow we on from 9:00 supposed to go from 9 to half past 3 with of course a couple of breaks in between. Um we supposed to be doing the 2023 paper. Saturday we have um a double hopefully have a double 10 to 12 and then 1 to 3. Um again working the 2024 paper and then Sunday evening we from 4 to about about 8 half 8 somewhere working the 2025 paper. So, of course, if if we finish a bit earlier, we could answer any extra questions or whatnot.
Or we could just stop on your Google.
All right. Okay. I'll log out there.
I'll see you.
>> See you later.
>> Yeah, later.
All right. Sorry, man. Yeah, boy. Jesse, I wish I was doing math, too. Let's I just do the scope capacity for that right now.
561.
>> No, we just finished.
I need to check another thing.
Oh, all right.
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