1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: (cosx)^2 half angle formula

  1. Feb 7, 2009 #1
    y=sin2x bounded by x=0,x=pi,y=0 , revolved around the x axis

    cross section A=pi(sin2x)^2


    taking u = sinx ; du=cosxdx

    im unclear on how to proceed in this case where du needs to satisfy (cosx)^2
    the problem hints to use a half angle formula
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 7, 2009 #2
    Re: sin2x,bound{x=0,x=pi,y=0}

    Try the identity cos(2x) = 1 - 2sin2(x). It can be rearranged to be a half-angle formula. :smile:
  4. Feb 7, 2009 #3
    Re: sin2x,bound{x=0,x=pi,y=0}

    still i am unclear on how to proceed here
  5. Feb 7, 2009 #4
    Re: sin2x,bound{x=0,x=pi,y=0}

    Try using the Pythagorean identity to get sin2(x) - sin4(x), then apply the half angle formula to get cosines (twice for the second term) that are not squared.
  6. Feb 7, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    Re: sin2x,bound{x=0,x=pi,y=0}

    slider142 is trying to say that since sin^2(x)=(1-cos(2x))/2, sin^(2x)=(1-cos(4x))/2. That's pretty easy to integrate.
  7. Feb 8, 2009 #6
    Re: sin2x,bound{x=0,x=pi,y=0}

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook