Hey, everyone. Now that we know how to simplify trig expressions, you're going to come across a new type of problem in which you're given a trig equation and asked to verify the identity. This sounds like it could be complicated, but here I'm going to show you that verifying an identity just comes right back down to simplifying, just with the specific goal of both sides of our equation being equal to each other. So here, we're going to keep using our simplifying strategies, just in a slightly different context. And I'm going to walk you through exactly how to do that here. So let's go ahead and get started. In working with these problems, you may only have to simplify one side of your equation or you may have to simplify both sides. This will become more apparent as you begin to work through a problem. Let's go ahead and jump right into our first example here where we're asked to verify the identity. Something that I do want to mention is that sometimes these problems may ask you to prove the identity or establish the identity, but these all mean the same thing.
The identity that we're asked to verify here is sinθcosθ 1−cos2θ is equal to 1tanθ. So how do we start here? Well, in working with these problems, we always want to start by simplifying our more complicated side first. So in looking at this equation, it's clear that this left side is more complicated than that right side. So we're going to start by applying our simplifying strategies to that left side. Looking at my left side, sinθcosθ 1−cos2θ , remember, one of our most important strategies for simplifying is to constantly be scanning for identities. Looking at this left side of my equation this one minus cosine squared of theta looks familiar. And looking at my identities here I know that one of my Pythagorean identities tells me that sin2θ+cos2θ=1. So if I subtract the cosine squared of theta from both sides of this identity, I end up with exactly what's in that denominator on that right side, 1−cos2θ. So using this identity, I can replace that denominator with sin2θ using that Pythagorean identity. Now I'm going to keep my numerator the same as sinθcosθ. And I can do some canceling here. Because in my denominator I have sin2θ and in my numerator I have sinθ. So that sinθ in my numerator goes away. And in my denominator my sin is no longer squared. And I'm simply left with cosθsinθ. Now remember we are still constantly scanning for identities cotθ. Now I also know that the cotangent is equal to 1tanθ, which is exactly what the right side of my equation is. And remember that our ultimate goal is to show that these two sides of our equation are equal to each other. So knowing that this cosine over sine is 1tanθ, I can go ahead and rewrite that side as 1tanθ. Now I don't have to do anything to that right side of my equation. It's just 1tanθ. And now we have that 1tanθ is equal to 1tanθ. So we have successfully verified this identity by showing that the left side of our equation is equal to the right side. Now, even though this is called verifying the identity, this is not a new identity you have to learn. All that we're doing here is showing that two sides of an equation are equal.
Let's go ahead and take a look at our second example here. We're still verifying the identity, but the trig equation that we're given here is sec2θ−tan2θ cos(−θ)+1 is equal to 1−cosθ sin2θ . Now remember that we want to start with our more complicated side first, and sometimes it won't be immediately apparent to you which side is more complicated. That's totally okay. Here, I'm going to start with the left side because this looks like there's a little bit more going on here. But if you were to start with the right side, you could still end up being able to verify this identity. But let's go ahead and get started in simplifying that left side, sec2θ−tan2θ cos(−θ)+1 . Remember here we want to be constantly scanning for identities. And looking at this particular side of my equation, this sec2θ minus tan2θ reminds me of a Pythagorean identity, which tells us that tan2θ+1=sec2θ. So if I rearrange this identity and subtract the tangent squared of theta from both sides, I will end up with exactly what's in that numerator there. And it's simply equal to 1. So I can replace that entire entire numerator with just 1. Then in my denominator, I still have that cos(−θ)+1. Now remember, we're still constantly scanning for identities here. And looking in that denominator, since I have the cosine of negative theta, I know that I can use my even-odd identity in order to get rid of that negative argument here because the cosine of negative theta is simply the cosine of theta. So simplifying that denominator further, leaving my numerator as 1 because that cannot be further simplified, in my denominator, I now just have the cosθ+1. Now this looks rather simplified. So let's go ahead and leave this as is and see if we can get this right side of our equation to be equal to that left side. So let's go ahead and start simplifying here. Here I have 1−cosθ sin2θ . Now, one of my simplifying strategies tells me that if I have one plus or minus a trig function, I want to go ahead and multiply the top and the bottom by 1−+cosθ or plus that same trig function. Here in my numerator, since I have 1−cosθ, I'm going to go ahead and multiply this by 1+cosθ. And remember, if I'm multiplying the top by that, I need to multiply the bottom by that as well to not change the value of our expression. So let's go ahead and multiply this out. Now in my numerator, that is a difference of squares. So this ends up giving me 1−cos2θ. Then in my denominator, I'm going to keep that as is, sin2θ times 1+cosθ. Here in my numerator scanning for identities, this one minus cosine squared of theta is an identity that we've used before, our Pythagorean identity. Now I know that one minus the cosine squared of theta is simply equal to sin2θ, so I can go ahead and rewrite that numerator as the sin2θ. And then in my denominator, I also have a sin2θ, and that is still multiplied by 1+cosθ. Now here I can cancel some stuff out because I have sin2θ on the top and on the bottom. Now I am simply left with a 1 in that numerator and then a 1+cosθ in that denominator, which is exactly what I have on that left side there. Now I can go ahead and rewrite that left side, just switching the 1 and the cosine, to get them exactly the same, 1+cosθ. And now my two sides of my equation are equal to each other, and we have successfully verified this identity.
Now I know that working through these sorts of problems can be really tricky, so if you're working through one and you just get stuck and you're not quite sure where to go, it can sometimes be a good idea to just completely start over from scratch because you may discover something that you didn't discover before. Just keep getting repetition with these problems and you'll get better and better with every try. Thanks for watching, and I'll see you in the next one.