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Yet another trig identity .

  1. Sep 11, 2008 #1
    yet another trig identity.....

    1. The problem statement, all variables and given/known data

    prove: cos^(x)= 5/16+15/32(cos2x)+3/16(cos4x)+1/32(cos6x)

    2. Relevant equations



    3. The attempt at a solution

    i attempted to use the formula cos^2(x)=(1+cos2x)/(2), and square both sides, then use it again for the square roots, then multiply the answer by (1+cos2x)/(2) again thus making the left side cos^6.........................not getting the right answer
     
  2. jcsd
  3. Sep 11, 2008 #2
    Re: yet another trig identity.....

    i think im on the right trrack?
     
  4. Sep 11, 2008 #3

    Defennder

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    Re: yet another trig identity.....

    You mean cos^2 (x) on the left?
     
  5. Sep 11, 2008 #4
    Re: yet another trig identity.....

    hey....the original proof is cos^6(x) on the left side
     
  6. Sep 11, 2008 #5
    Re: yet another trig identity.....

    i get to a certain point where i get like....cos^2(4x) times cos (2x) wth a bunch of other stuff on the left....but i dont knwo what to do with it
     
  7. Sep 11, 2008 #6

    Defennder

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    Re: yet another trig identity.....

    Oh yeah, should have read it more closely. I haven't tried it out yet, but if you know the left is function of cos x only, you should repeatedly apply the double angle cos and sin formula to change everything to functions of sin x and cos x. Then express those functions of sin x as functions of cos x. It should all cancel out.
     
  8. Sep 11, 2008 #7
    Re: yet another trig identity.....

    lol its cool.....so wadda u think? do you get what i tried to do? like is it clear when i explained it?
     
  9. Sep 12, 2008 #8
    Re: yet another trig identity.....

    thats what i keep doing...
     
  10. Sep 12, 2008 #9

    Defennder

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    Re: yet another trig identity.....

    I just proved it using the approach suggested earlier. Just convert everything on the right to a function of cos x. A lot of terms will cancel out to give cos^6 x.
     
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