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Thread: Finding the exact value

  1. June 10th 2008, 07:51 PM #1
    gabrie30
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    Finding the exact value

    of this expression is driving me crazy...

    sin(2sin^(-1)(-2/5))




    anyone willing to enlighten me?
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  2. June 10th 2008, 08:31 PM #2
    mr fantastic
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    Quote Originally Posted by gabrie30 View Post
    of this expression is driving me crazy...

    sin(2sin^(-1)(-2/5))




    anyone willing to enlighten me?
    Let \alpha = \sin^{-1} \left( - \frac{2}{5} \right) \Rightarrow \sin \alpha = - \frac{2}{5}.

    You want the value of \sin (2 \alpha) = 2 \sin \alpha \cos \alpha.

    I will assume you can find \cos \alpha given \sin \alpha = - \frac{2}{5} ......
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