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Thread: Help with trig functions please

  1. #1
    Junior Member
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    Help with trig functions please

    I need to know how

    $\displaystyle 1-\cos(2\arcsin\frac{s}{4}) = \frac{s^{2}}{8} $

    and how

    $\displaystyle \sin(2\arcsin\frac{s}{4}) = 2\frac{s}{4}\sqrt{1-\frac{s^{2}}{16}}$

    I know the rule $\displaystyle \cos\arcsin(x)) = \sqrt{1-x^{2}} $ but don't know what to do with the 2 before the inverse trig functions eah time.

    Thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by bobby View Post
    I need to know how

    $\displaystyle 1-\cos(2\arcsin\frac{s}{4}) = \frac{s^{2}}{8} $
    You need the identity $\displaystyle \cos(2\theta)= \cos^2(\theta)- \sin^2(\theta)$ and replacing $\displaystyle \cos^2(\theta)$ by $\displaystyle 1- \sin^2(\theta)$, $\displaystyle \cos(2\theta)= 1- 2\sin^2(\theta)$.

    So $\displaystyle 1- \cos(2\arcsin(\frac{s}{4})= 1- 1+ 2\sin^2(\arcsin(\frac{s}{4}))$$\displaystyle = 2s\\in^2(\arcsin(\frac{s}{4}))$.

    Surely you know what $\displaystyle sin(\arcsin(\frac{s}{4})$ is?

    and how

    $\displaystyle \sin(2\arcsin\frac{s}{4}) = 2\frac{s}{4}\sqrt{1-\frac{s^{2}}{16}}$
    $\displaystyle \sin(2\theta)= 2 \sin(\theta)cos(\theta)$

    I know the rule $\displaystyle \cos\arcsin(x)) = \sqrt{1-x^{2}} $ but don't know what to do with the 2 before the inverse trig functions eah time.

    Thanks
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  3. #3
    Junior Member
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    Thanks very much HallsofIvy
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