Taylor Series (Derivative question)

[newbe318]

I was looking at the solution for problem 6 and I am confused on taking the derivatives of the function f(x)= cos^2 (x)
I took the first derivative and did get the answer f^(1) (x)= 2(cos(x)) (-sin (x)), but how does that simplify to -sin (2x)?

Is there some trig identity that I am not aware of?
Your help would be much appreciated, since I have a final Friday.
Thank you.

 

[Mark44]

View attachment 93145

I was looking at the solution for problem 6 and I am confused on taking the derivatives of the function f(x)= cos^2 (x)
I took the first derivative and did get the answer f^(1) (x)= 2(cos(x)) (-sin (x)), but how does that simplify to -sin (2x)?

Is there some trig identity that I am not aware of?

This is a well-known trig identity.
$\sin(2x) = 2\sin(x)\cos(x)$
Another is $\cos(2x) = \cos^2(x) - \sin^2(x)$
The right side can be written in two other forms: $2\cos^2(x) - 1$ or $1 - 2\sin^2(x)$.

Your help would be much appreciated, since I have a final Friday.
Thank you.