Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor Series (Derivative question)

  1. Dec 9, 2015 #1
    Screen Shot 2015-12-09 at 11.12.57 AM.png

    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.
     
  2. jcsd
  3. Dec 9, 2015 #2

    Mark44

    Staff: Mentor

    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)##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Taylor Series (Derivative question)
Loading...