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Simply Trig Expression

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Simplify the expression:

    cos(2sin^-1 (5x))

    2. Relevant equations

    Fundamental identities: 1 = sin^2 ϑ + cos^2 ϑ : I think you use this one?

    3. The attempt at a solution

    Let y=2sin^-1(5x)
    sin y = 10x

    so, you plug in?
    1 = 10x^2 + cos^2 y

    not really sure if im on the right path or what to do next
     
  2. jcsd
  3. Sep 25, 2007 #2

    Avodyne

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    Do you know a "double angle formula" that expresses cos(2t) in terms of sin(t)?
     
  4. Sep 25, 2007 #3
    cos 2t = 1 - 2sin²t

    let t = sin^-1 5x
    so sin t = 5x

    cos 2t = 1 - 2(5x²)
    cos t = ( 1 - 2(5x²) ) / (2)

    is this correct?
     
    Last edited: Sep 26, 2007
  5. Sep 26, 2007 #4

    Avodyne

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    cos(2t)=1-2(5x2) is almost correct; if t=arcsin(5x), what is sin2t ? But your last line is not correct. (Can you see why?) However, you don't need the last line; you have simplified the expression, and you're done!
     
  6. Sep 26, 2007 #5
    sin²t = 5x²?
     
  7. Sep 26, 2007 #6
    is cos 2t = 1 - 2(5x²) in simplest form? is that the answer?
     
  8. Sep 26, 2007 #7

    Avodyne

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    sin t = 5x, so sin2t=(5x)2=25x2.

    But I'm sure you knew that ...
     
  9. Sep 26, 2007 #8
    Oh duh, ok from the start:

    cos(2 arcsin 5x)
    Let t = arcsin 5x
    so, sin t = 5x

    Since cos 2t = 1 - 2sin²t

    cos 2t = 1 - 2(5x)²
    cos 2t = 1 - 2(25x²)
     
  10. Sep 26, 2007 #9

    Avodyne

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    That's it. You might want to simplify it further to 1-50x2, but that's a minor detail.
     
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