In this video, we're going to be going through 14 verifying trigonometric identity problems. I want to challenge you to see how many of these can you get right on your own. Put the answer down in the comment section below.
Let's jump into the first one. We've got cosecant squared theta minus one divided by cosecant squared theta, and we want to show or prove or verify that it equals cosine squared theta. So the first thing you want to do when you tackle these identities is you want to look at which side is like larger or more expanded or complicated and then we'll work with that side and we'll condense it down to the side that's more condensed or compact.
So in this case it clearly seems to be the left side that's more expanded. So let's start there. We've got cosecant squared theta minus one.
Whenever I see the squared trig functions I always think of these three Pythagorean trig identities right here. It doesn't always have to be these but oftentimes it's helpful to use these. So here we have cosecant squared If we subtract 1 from both sides, see how cosecant squared minus 1 equals cotangent squared? So we can go ahead and make that replacement. So we're going to replace cosecant squared theta minus 1 with cotangent squared.
Okay, and this is all over cosecant squared theta. And we're trying to verify or prove that it equals cosine squared theta. So the next thing I want to do is when I look at cotangent squared, you can see that cotangent equals cosine over sine. But if cotangent is squared, then we're going to square this side as well.
It's going to be cosine squared over sine squared. So let's go ahead and replace that numerator with cosine squared theta over sine squared theta. And then cosecant squared theta, we know that cosecant is 1 over sine.
But if this is squared, this will be 1 over sine squared. So let's replace that. So when you're thinking about identities, really, you know, these quantities, they're identical.
We're just interchanging one. with the other. So we're not really changing the problem, we're just changing the way that it looks. Now here what we're going to do is we've got cosine squared over sine squared, one over sine squared.
Let's go ahead and multiply the numerator and the denominator by sine squared theta because anything divided by itself, this is one, right? And so what happens is the sine squares are going to cancel and we've got cosine squared over one which equals cosine squared theta, which is what we were trying to prove that it's equal to, and you got it. Okay, number two, see if you can do this one on your own.
Remember when you were in geometry and you had to do those two-column proofs? This is kind of what this is like. It's just that we're doing it with trig functions.
So here, when we look at the left side and the right side, which one do you think is more expanded? Well, definitely the right side. You can see we've got two fractions here, and here we just have one term here. So what we want to do is we want to combine these together into one term, right? The nice thing about these is that we know what the answer is.
That's the good news. The bad news is there's not one set, you know, way of doing each one of these problems. But as long as you make an identical substitution, you know, you're going to have the same quantity.
It's just in a different form. So definitely want to practice a lot of these to get good at them. But here, the first thing I notice is that, you know, just like in algebra, you know how you had to get a common denominator? We want to get a common denominator for both these fractions.
This one has a 1 minus cosine theta. But it doesn't have a 1 plus cosine theta. And this has a 1 plus cosine theta, but it doesn't have a 1 minus cosine theta.
So what I'm going to do here is I'm going to multiply by what it's missing. So here I'm going to multiply this by 1 plus cosine theta. And remember, whatever you do to the denominator, you have to do that to the numerator. Okay, because that's like multiplying by 1. Here we're missing a 1 minus cosine theta. So we're going to multiply the numerator by 1 minus cosine theta.
Okay, so now let's go ahead and simplify a little bit. Here we have a binomial times a binomial. What you want to do is you want to multiply this together. So you have 1 times 1, which is 1. You have cosine theta and negative cosine theta, which cancel. And then we have cosine theta times negative cosine theta, which is negative cosine squared theta.
And what I'm going to do is when I combine these together, since they have the same denominator now, we're going to put it all over this common denominator. In the numerator, we have 1 plus cosine theta. And then over here we have one minus cosine theta. Okay, now keep in mind you want to keep one eye on the left side, one eye on the right side.
We know where we want to go to, so we want to make sure we're heading in that right direction. But at this point we're just trying to simplify or condense as much as we can. So here you can see I've got a cosine theta and a negative cosine theta. Those cancel. Here we've got one plus one, which is two, over 1 minus cosine squared theta now remember when I said you see these ones with the second degree the squared ones I always think of these three Pythagorean trig identities.
Well, that's what we have here We've got one if I subtract cosine squared theta from both sides of this equation It's still going to be an identity is still going to be equivalent. But now we have sine squared equals 1 minus Cosine squared. So let's replace this denominator with what it equals, which is sine squared theta. Okay, now we're almost there.
Notice the sine is in the denominator, right? So you can see that this is actually sine squared. So you can think of this as 2 times 1. Okay, 2 times 1 is still 2. But look, 1 over sine squared we know is going to be cosecant squared.
So this is going to equal 2 cosecant squared theta. And that's what we were trying to prove that it was equal to. And you got it. For three, see if you can do this one.
We've got cotangent squared theta plus one times sine squared theta minus one, and we're trying to show that it equals negative cotangent squared theta. So when I was learning this in school myself, you know, my teachers, they would say, if you don't know what to do, just do something. Just make some type of substitution.
If it's the wrong one, meaning, you know, maybe it gets more complicated, you can always back up a couple steps and start over and try a different substitution, but don't stay stuck. You definitely want to try. some type of substitution and try to move forward. You're going to get better as you do these more and more and you'll start to recognize what's going to be an easier substitution as you get experience. So in this one, again, we want to look at which side is more expanded and which side is more condensed, right?
So what do you think? It looks definitely like the left side is more expanded. So let's start there. Now notice we've got those second degree trig functions, right?
They're squared. Again, like we've been talking about, we use these Pythagorean trig identities a lot. It's good to memorize them or You know at least write them down and have them in front of you.
So let's take a look what we have here We've got cotangent squared theta plus 1 see cotangent squared squared theta plus 1 We know that equals cosecant squared theta. So let's replace that with cosecant squared theta Okay. Now over here. We have sine squared theta minus 1 now Follow me on this one see if I subtract 1 to the left side of the equation And if I subtract cosine squared theta to the right side sine squared theta minus one actually equals negative cosine squared theta. So we're going to replace this with negative cosine squared theta.
Okay, now let's take a look at what we have so far. It's a little bit more condensed. Cosecant squared, we know, is one over sine squared.
So let's go ahead and make that replacement, one over sine squared theta. And then over here we've got negative cosine squared theta. Okay. Let's just leave that as negative cosine squared theta. Another little hint is if you ever get stuck, all these trig functions are basically made up of sines and cosines.
So if you get really stuck, you can always convert back to sines and cosines. Those are the basic building blocks, and you can figure it out from there. But I try not to do that at the very beginning.
If I can avoid it, I try to use some of the other identities first. But that's always the last kind of bet that you can use. So we've got negative cosine squared theta over sine squared theta. And now let's take a look at what we have.
Cosine over sine is cotangent, but these are both squared. This is going to be cotangent squared because both sides are squared. And we have a negative, so it's going to be negative cotangent squared theta.
And you can see we've got a match, so we proved it. Okay, number four, see if you can do this one. Cosecant theta plus cotangent theta equals sine of theta over 1 minus cosine theta. How would you do that one?
Well, first of all, I wanted to say that, you know, congratulations. Most people probably have bailed already on this video because it seems too challenging, but I want to congratulate you for sticking with it. It definitely, the more you do, the better you're going to get at it.
And, you know, I'm here to cheer you on. You know, that's what my whole channel is really devoted to. You know, in athletics, you know how they have cheerleaders that are cheering on the athletes. Well, you guys are mathletes, right? And so what we want to do here is we want to improve our math skills.
So in this one, what do you think is the better side to work with? You really can't make a mistake, you know, but in this one I'm going to start with the right side and we're going to try and work with that first. So here what I'm going to do is I'm going to use a little technique called multiplying by the conjugate. So conjugate is you change the sign in between these two terms of this binomial.
So I'm going to multiply by 1 plus cosine theta and 1 plus cosine theta. And you're probably saying, Mario, how did you know how to do that, right? Part of it comes with experience. When you see these come up over and over again, you'll recognize this as a...
A valid technique. So let's go ahead and multiply the denominators together first. We've got 1 times 1, which is 1. Again, the inside and outside cancel.
We have negative cosine theta and positive cosine theta. And then the last terms give us negative cosine squared theta. In the numerator, I'm just going to leave this factored as sine theta and 1 plus cosine theta because sometimes what happens is you're going to be able to get some terms that cancel in the numerator and denominator.
When I look at 1 minus cosine squared theta, I see that squared, so I'm thinking Pythagorean trig identities possibly, right? 1 minus cosine squared theta will leave us with sine squared theta. So I'm going to replace this denominator with sine squared theta.
I'm going to leave the numerator the same, sine theta times 1 plus cosine theta. Now notice we've got 1 sine theta. Here, 2 sine, you know, sine squared theta here, basically sine times sine.
So what you can do is you can think of one of these sines canceling with one of these sines, and we just have sine theta left over in the denominator. So now let's see what we have left. We've got 1 plus cosine theta over sine theta. What I'm going to do here is I'm going to actually split this up into two fractions.
I'm going to say 1 over sine theta plus cosine theta over sine theta. See, I'm just like splitting it up into two parts. So let's write that down. 1 over sine theta.
plus cosine theta over sine theta. And 1 over sine theta, we know, is cosecant theta, right? Right here. So that's going to be cosecant theta.
These are called the reciprocal identities. And cosine over sine theta, we know that that is cotangent theta. And you can see we've got a match.
So we verified or we proved it. Okay number five, how would you do this one? We've got cotangent to the fourth theta equals cotangent squared theta times cosecant squared theta minus cotangent squared theta.
Okay so that was a mouthful. So I think here I'll start with the right side. It looks like it's a little bit more expanded. And the technique I'm going to use here on this one is I'm going to actually see this group here and see this group here.
See how they have a cotangent squared theta in common? Let's go ahead and factor that out as the greatest common factor. So cotangent squared theta, and then we're left with cosecant squared theta minus 1. The nice thing about factoring, you can check your work. If you distribute back in, you should get back to the original. Now notice how we have this cosecant squared theta.
When I have this squared, I, again, think of the Pythagorean trig identities. Cosecant squared minus 1 would leave us with cotangent squared on the left there. So let's go ahead and replace this with cotangent squared theta. And this is multiplied.
by cotangent squared theta. And we've got cotangent times cotangent. When you multiply, just like with an algebra, right, the rules of exponents, you add the exponents. So it's going to be cotangent to the fourth theta, which is what we have here, and we got it.
So some of the problems that we're doing are a little bit more challenging. Some of them are a little bit more easier. I tried to get a nice mix in here, but I'd say they're... you know kind of in the middle at about the medium level of difficulty. Number six how would you do this one we've got one plus cosine theta over sine theta plus sine theta over one plus cosine theta.
equals two cosecant theta. So definitely the left side seems more expanded. Let's start with that one. The key here would be really to combine these into one fraction with a common denominator.
You can see there's like two terms here. There's only one term here. So what's the common denominator? Well, we have to multiply by what's missing, right?
So here you've got one plus cosine theta. We need a sine theta. So we're going to multiply top and bottom by sine theta.
So that's going to make this sine squared theta. Okay, and it's all over the common denominator, which is going to be sine theta, 1 plus cosine theta. Over here, we've got a sine theta, but we're missing a 1 plus cosine theta.
So we're going to have to multiply 1 plus cosine theta times 1 plus cosine theta. which is 1 plus cosine theta, the quantity squared, okay, because the same thing twice. And this is added.
So now let's go ahead and simplify this out. If we FOIL this out, 1 plus cosine theta times 1 plus cosine theta, we get 1 plus 2 cosine theta plus cosine squared theta plus sine squared theta all over our common denominator, sine theta times 1 plus cosine theta. Now, cosine squared plus sine squared... That's your basic Pythagorean trigonometry. That's going to equal 1. So we can replace this here with 1. And notice we also have 1 and 1 here.
So let's go ahead and combine those together to 2. So we've got 2 plus 2 cosine theta over sine theta 1 plus cosine theta. Now notice we can factor out a 2 here in the numerator. So this is 2 times 1 plus cosine theta all divided by 1. sine theta, one plus cosine theta.
And you see the numerator and denominator, that one plus cosine theta cancels. We have two over sine theta, which is equal to two cosecant theta, and we've got a match, and you've proved it. Okay, let's try number seven and number eight.
These ones are a little bit easier. We've got cotangent of pi over two minus theta times cotangent of theta, and we wanna show that it equals one. When you see this pi over two minus theta, or 90 minus theta, these are called cofunctions.
Now let's go down here to the bottom. It's co-functions. Cotangent of pi over 2 minus theta is equal to tangent of theta.
So let's go ahead and replace this with tangent of theta times the cotangent of theta. Now we know that cotangent and tangent, they're reciprocals of one another. So we can replace cotangent with 1 over tangent. And then tangent of theta, we can just think of that as tangent of theta over 1. And these are going to cancel on the diagonal there, numerator and denominator.
We get 1 over 1, which is 1. and we've proved it, one equals one. Okay, try number eight now. Negative sine of negative theta divided by negative cosine of negative theta. We wanna show that.
equals negative tangent theta. Now when you see these negative angles, these are what are called even and odd identities, and that's these ones right here. So the sine of negative theta is the same as sine of positive theta, but it's multiplied by negative one. Okay, this is called an odd function because it's the opposite, right? But we have this negative here, so we've got negative, sine of negative theta we set as negative sine positive theta, okay, all divided by negative cosine of negative theta.
So cosine of negative theta is the same as cosine of positive theta. So this is going to be cosine of positive theta multiplied by this negative here in front. And now you can see a negative times a negative, that's going to be a positive. A positive divided by a negative is going to be a negative.
And sine divided by cosine is tangent. So we've got negative tangent of theta. equals negative tangent of theta and we proved it. Okay, number nine, see if you can do this one.
You've got cosecant to the fourth theta times cosecant theta cotangent theta minus cosecant squared theta times cosecant theta cotangent theta and we wanna show it equals cosecant cubed theta times cotangent cubed theta. So this one's a real large one. Definitely the left side is more expanded than the right side. And what I'm gonna do on this one is Factor out the greatest common factor. So you can see that this whole thing is a group.
This whole thing is a group. We want to see what they have in common. Let's start off with what's in front here. We've got cosecant squared theta cosecant to the fourth theta. So we can factor out a cosecant squared theta.
Also look what's in parentheses. You've got a cosecant theta cotangent theta. So let's also factor out a cosecant theta cotangent theta.
And if we do that, what we're gonna be left with is cosecant squared cosecant squared theta minus one. Okay, because we factored out this whole thing here, so we're just left with one. Here we factored out cosecant squared, so we were just left with cosecant squared when we divided this out of here.
So now look what we've got. We've got cosecant squared theta minus one. If we subtract one from both sides of this Pythagorean trig identity, you can see we're gonna be left with cotangent squared. theta. And now let's see what we've got.
We've got cosecant squared times cosecant to the first. That's going to give us cosecant cubed. Here we have cotangent to the first times cotangent to the second.
That gives us cotangent cubed. And we've got it. So that was an easy one once you factored out the greatest common factor. Number 10, how would you do this one?
1 plus cosine theta times 1 minus cosine of negative theta. We want to show that equals sine squared theta. So definitely easier to start with on the left side. It's more expanded. You can see we've got a negative angle, so we're thinking of our even and odd identities.
The cosine of negative theta we know is the same as cosine of positive theta. Okay, so this whole thing is going to be cosine positive theta, and we still have 1 minus. And this over here is 1 plus cosine theta.
And now if we FOIL this together, 1 times 1 is 1. cosine theta and negative cosine theta, those cancel, and then cosine theta times negative cosine theta is negative cosine squared theta. You can see we've got that squared cosine, so we can think of our Pythagorean trig identities. 1 minus cosine squared equals sine squared, and we proved it. Okay, number 11, how would you do this one? Sine theta times tangent theta all over 1 minus cosine theta minus 1. equals secant theta.
Alright, so on this one, you can see the left side's more expanded. We want to condense it down to secant theta. But what I have here is I've got two groups, two terms separated by this minus sign.
So what we really want to do is combine them into one group by getting a common denominator. But let's say our common denominator is going to be 1 minus cosine theta. So I'm going to multiply the numerator and denominator here by 1 minus cosine theta. So that gives us 1 minus cosine theta over 1 minus cosine theta. Okay, and because we had a common denominator, we're just going to extend this fraction here, sine theta, and tangent theta is sine theta over cosine theta, right here, tangent theta.
So now what I'm going to do is... I'm going to do a little bit of simplifying here. I'm going to distribute the negative.
And sine theta is like sine theta over 1. So I'm going to multiply the numerators and denominators. So I've got sine squared theta over 1 times cosine is cosine theta minus 1. 1, and then a negative times a negative gives you plus cosine theta, all over the common denominator 1 minus cosine theta. And keep in mind, we're trying to show that this equals secant theta. So now what's interesting is we have this fraction within a larger fraction.
So what we want to do is we want to get rid of that complex fraction. I'm going to do that by multiplying the numerator by cosine theta and the denominator by cosine theta, because that's like multiplying by 1. So if I distribute cosine to each of these terms in the numerator. Here the cosines are gonna cancel and that's just gonna leave us with sine squared theta. Cosine theta times negative one is negative cosine theta. Cosine theta times cosine theta is cosine squared theta.
And then the denominator, I'm just gonna leave this as cosine theta times one minus cosine theta. And the reason I'm leaving it factor is because we might get some cancellation numerator and denominator. Now notice how we have sine squared. plus cosine squared and that's our basic Pythagorean trig identity that equals one so this gives us one minus cosine theta over cosine theta times one minus cosine theta so you can see that this one minus cosine theta and the numerator and denominator cancel out and we just have one over cosine theta which equals secant theta which is what we were trying to prove okay see if you can do number 12 and 13 here we've got cosecant squared theta minus tangent squared of pi over 2 minus theta and we'll try to show that equals 1 so definitely the left side is more expanded notice we've got this pi over 2 minus theta we want to refer to our co-function identities and you can see over here a tangent of pi over 2 minus theta is equal to cotangent of theta but because this is tangent squared then this is going to be cotangent squared it's like we're squaring both sides so this is going to be cotangent squared theta Okay, cotangent squared theta, and this is cosecant squared theta minus cotangent squared theta. And again, when you have these ones to the second power, I like to think of the Pythagorean trig identities.
Cosecant squared, if we subtract the cotangent squared over, that's going to leave us with 1. So you can see 1 equals 1. That was kind of a quick and short one. Let's try number 13 now. 1 over sine theta plus 1 plus 1 over sine theta minus 1. We want to show that equals negative 2 tangent theta secant theta.
All right, how would you do that one? Well, notice how we have one group here. We've got two groups here.
We really want to combine these into one fraction by getting a common denominator. So what I'm going to do is I'm going to multiply by what I'm missing. I've got a sine theta plus one, but I need a sine theta minus one.
So I'm going to multiply the numerator and denominator by sine theta minus one. So that's going to make this sine theta minus one over our common denominator, sine theta plus one times sine theta minus one. which when you multiply that out, it comes out to sine squared theta minus 1. And then here we're missing a sine theta plus 1, so I'm going to multiply the numerator and denominator by sine theta plus 1 all over our common denominator, sine squared theta minus 1. Now you can see the negative 1 and the positive 1 cancel. Sine theta plus sine theta is 2 sine theta over sine squared theta minus 1. Now what's sine squared theta minus 1 equal to? Well, if we go to our Pythagorean tree identities, if we subtract 1 to the left and subtract the cosine squared to the right, sine squared minus 1 equals negative cosine squared.
So now we have 2 sine theta over negative cosine squared theta. Okay, so how are we going to work with this one? Well, what you can do is you can split this up. You can say this is 2 sine theta over, now cosine squared is really like, let's say, 1 sine theta over 2 sine squared. You've got a cosine theta times another cosine theta.
Okay, so see how that gives you back the negative cosine squared theta? But one over cosine theta equals secant theta, and sine over cosine is tangent, and two over negative one is negative two. So you proved it.
Last one, number 14. Remember to put down in the comments below how many of these you were able to get right. Let's see if you can solve this one. Tangent of theta over cosecant theta. We want to show it equals secant theta minus cosine theta. Now remember, there's no right or wrong way to do these.
You just want to stick with one side. If you decide you're going to work with the right, show it equals the left, or start with the left, show it equals the right. Either way, I'm going to start with the left on this one.
Tangent of theta is equal to sine of theta over cosine theta. Okay, and cosecant theta is 1 over sine theta. Now remember, when you divide by a fraction, It's like multiplying by the reciprocal.
So if we take this 1 over sine and flip it and multiply it by the numerator, that's going to give us sine squared theta over cosine theta. Now sine squared theta, if we go to our Pythagorean trig identity here, we can subtract the sine squared to the other side. Oops, I said that wrong. We're going to subtract the cosine squared to the other side.
Sine squared is equal to 1 minus cosine squared. So we're going to replace sine squared with 1 minus cosine squared, all divided by cosine theta. Now... what I can do is I can split this up into two fractions, one over cosine theta, which is equal to secant theta, and I can split this up over here minus cosine squared divided by cosine is just going to give us cosine theta, and we proved it.
If you want to see more examples, follow me over to that video right there. We'll do some more trig identities. I'll see you in the next video.