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Math Help - inverse trig functions

  1. #1
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    Lightbulb inverse trig functions

    im clueless where to begin...

    Simplify 8|cos(θ)| if θ = sin^-1(x/8) for some real number x.

    soo....

    8|cos(sin^-1(x/8)|

    than...

    8|sq. root(1-sin^2(sin^-1(x/8))|

    i got lost there.....
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  2. #2
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    Hello, r0b0tix!

    \text{If }\theta\:=\:\sin^{\text{-}1}\!\left(\frac{x}{8}\right),\:\text{ find: }\:8|\cos\theta|

    We have: . \theta \:=\:\sin^{\text{-}1}\!\left(\frac{x}{8}\right) \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{8}

    \text{Then: }\:\cos^2\!\theta \;=\;1-\sin^2\!\theta \;=\;1-\left(\frac{x}{8}\right)^2 \;=\;\frac{64-x^2}{64}

    . . \text{Hence: }\:\cos\theta \;=\;\pm\frac{\sqrt{64-x^2}}{8}


    \text{Therefore: }\:8|\cos\theta| \;=\;8\left|\pm\frac{\sqrt{64-x^2}}{8}\right| \;=\;8\cdot\frac{\sqrt{64-x^2}}{8} \;=\;\sqrt{64-x^2}

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