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Math Help - [SOLVED] Evaluate cos(arcsin(t))?

  1. #1
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    [SOLVED] Evaluate cos(arcsin(t))?

    What do they mean exactly by evaluate cos(sin^-1(t))?
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  2. #2
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    Draw a triangle diagram, or cheat by consulting

    Compositions of Trig and Inverse Trig Functions

    ... assuming evaluate is meant to mean simplify.
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    Hmmm we haven't done much of this in class. Usually when a sin or cos pops up it's to calculate the rate of change or find the derivative. Any clues on how to simplify this? I looked at the link and don't really recognize doing anything similar in class.
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    Quote Originally Posted by thekrown View Post
    Hmmm we haven't done much of this in class. Usually when a sin or cos pops up it's to calculate the rate of change or find the derivative. Any clues on how to simplify this? I looked at the link and don't really recognize doing anything similar in class.
    Yes, but this particular task isn't actually calculus, just trig and pythag.

    Draw a right triangle. Label one of the two 'legs' as t = sin a. Label one of the corners (which one?) as a = arcsin t. Label the remaining leg as cos a = cos(arcsin(t)) = ... using pythag, sqrt(1 - t^2).
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by thekrown View Post
    What do they mean exactly by evaluate cos(sin^-1(t))?
    Let y=\arcsin(x) then y'=\frac{1}{\sqrt{1-x^2}}. But, we could have also done that y=\arcsin(x)\implies \sin(y)=x\implies y'\cos(y)=1\implies y'=\frac{1}{\cos(y)}=\frac{1}{\cos(\arcsin(x))}. Comparing these two gives \cos(\arcsin(x))=\sqrt{1-x^2}
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  6. #6
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    f(t)=\cos(\arcsin t) makes sense as long as |t|\le1, now f(t)\ge0, so by using \cos t=\pm\sqrt{1-\sin^2t} and putting t\mapsto\arcsin t we get f(t)=\pm\sqrt{1-t^2}, but f\ge0, thus f(t)=\sqrt{1-t^2} and we're done.
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  7. #7
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    Hmm I don't really get this to much. Thanks for your help guys but I think I'm going to need some tutor assistance with this.
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