Hey everyone. In this problem, we're asked to use our double angle formulas in order to verify the identity. The cotangent of 2θ is equal to the cotangent of θ minus 1 over 2 cosine θ sine θ. Now you should go ahead and try this on your own before checking back in with me. And remember that there are always multiple ways to verify an identity. So, if you did this slightly differently than me but you still ended up with both sides being equal to each other, that's totally fine. Now here, let's go ahead and jump in.
Remember, when verifying an identity, we want to start with our more complicated side first. And here, this right side is clearly more complicated, so let's start there. Using our simplifying strategies, one thing that I notice is that one of my terms is in terms of cosine and sine and the other one is not. So, let's go ahead and break down everything in terms of that sine and cosine. Now the cotangent of θ is equal to the cosine of θ over the sine of θ. So, I'm going to go ahead and replace that cotangent using that identity. Now, I'm still subtracting 1 over 2 times the cosine of θ times the sine of θ.
Now where do we go from here? We want to go ahead and combine any fractions using a common denominator since we have 2 separate fractions here. Now, here, my common denominator is going to be 2 times the cosine θ times the sine θ by multiplying this fraction by 2 cosine θ on the top and at the bottom. Now what happens when I do that? Well, I have 2 cosine2θ minus 1 over that common denominator, 2 cosine θ sine θ. Now let's go ahead and do some scanning for identities. In that numerator, I see 2 cosine2θ minus 1, which I recognize from my double angle identity for cosine here.
Now I can replace that with cos2θ.
But what about my denominator? Well, in my denominator, I see 2 cosine θ sine θ, which is exactly what the sine of 2θ is. So that becomes my denominator here, sine of 2θ. Now let's remember what our goal is, that left side of our equation, the cotangent of 2θ. Our goal is that this right side is equal to this left side and look at where we are. We have the cosine of 2θ over the sine of 2θ. And what is that equal to? It's equal to the cotangent of 2θ, which is exactly what that left side already is. So, we have successfully verified this identity having our right and left side of our equation being equal to each other. Thanks for watching, and let me know if you have questions.