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Math Help - cos((sinx)^-1)

  1. #1
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    cos((sinx)^-1)

    i was wondering if anyone could help me to simplify cos((sinx)^-1) i remember someone saying in class that it was something like
    2sqareroot(9-x^2) but i don't have a clue how to get there. any help would be much apreciated
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  2. #2
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    Let \theta = sin^{-1}(x). You can think of this as the angle whose sine is x/1. In a right triangle, sine is opposite over hypotenuse, so draw a right triangle, label one of the acute angles as theta, the opposite side as x and the hypotenuse as 1. Use the Pythagorean theorem to find the missing (adjacent) side:
    a^{2} + b^{2} = c^{2}
    x^{2} + b^{2} = 1^{2}
    Solve for b. Label this on the adjacent side.

    Now cos(\theta) = cos(sin^{-1}(x)), and cosine is adjacent over hypotenuse. Can you find cos(sin^{-1}(x)) now?


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  3. #3
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    Or... :

    \cos(t)=\pm \sqrt{1-\sin^2(t)} (from the well-known identity \cos^2(t)+\sin^2(t)=1

    But since the arcsin ranges from -\pi/2 to \pi/2, \cos(\arcsin(x))\geq 0

    And hence \cos(\arcsin(x))=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}
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