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. Now, 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. Now, in working with these problems, you may only have to simplify one side of your equation, or you may have to simplify both sides. And this is something that will become more apparent as you begin to work through a problem. So let's go ahead and jump right into our first example here where we're asked to, of course, verify the identity.
Now 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. Now, the identity that we're asked to verify here is: sin θ ⋅ cos θ 1 − cos 2 θ = 1 tan θ 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.
Now, looking at my left side, sin θ ⋅ cos θ ÷ ( 1 − cos 2 θ ), remember, one of our most important strategies for simplifying is to constantly be scanning for identities. And looking at this left side of my equation this 1 − cos 2θ looks familiar. And looking at my identities here, I know that one of my Pythagorean identities tells me that sin2 θ + cos2 θ = 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 minus the cos2 θ. So using this identity, I can replace that denominator with the sin2 θ using that Pythagorean identity. Now I'm going to keep my numerator the same as sin theta times cosine theta. And I can do some canceling here. Because in my denominator, I have sin2 θ and in my numerator, I have the sin of theta. So that sin theta in my numerator goes away. And in my denominator, my sin is no longer squared. And I'm simply left with the cosine of theta over the sine of theta. Now, remember we are still constantly scanning for identities cotangent of theta.
Now I also know that the cotangent is equal to 1 over the tangent, 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 1 over the tangent, 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 1 over the tangent of theta. And now we have that 1 over the tangent is equal to 1 over the tangent. 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 that you have to learn. All that we're doing here is showing that two sides of an equation are equal.
So 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: sec 2 θ − tan 2 θ cos ( − θ ) + 1 = 1 − cos θ sin 2 θ . 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. And that's totally okay. Now 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 by simplifying that left side, sec 2 θ − tan 2 θ ÷ cos ( − θ ) + 1 . Now remember here, we want to be constantly scanning for identities. And looking at this particular side of my equation, this secant squared minus the tangent squared reminds me of a Pythagorean identity, which tells us that tan2 θ + 1 is equal to sec2 θ. 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 numerator with just 1. Then, in my denominator, I still have that cosine of negative theta plus 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 cosine of theta plus 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 − cos θ ÷ sin2 θ . 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 minus or plus that same trig function. Now here in my numerator, since I have 1 minus the cosine of theta, I'm going to go ahead and multiply this by 1 plus the cosine of theta. And remember, if I'm multiplying the top by that, I need to multiply the bottom by that as well not to 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 − cos2 θ . Then in my denominator, I'm going to keep that as is, sin2 θ ⋅ 1 + cos θ . Now here in my numerator scanning for identities, this 1 − cos2 θ 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 sin2 θ, so I can go ahead and rewrite that numerator as the sin2 θ. And then in my denominator, I also have a sin<---