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Math Help - inverse function question.

  1. #1
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    inverse function question.

    For the problem:

    sin(arccos(-6/7))

    Would it be correct that I figure out arccos(-6/7) first and then plug that answer into sin? That's what I've been doing, but it's out of domain since the answer is around 2.6 when I plug the inverse of cos(-6/7) into the calculator.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by augustfai View Post
    For the problem:

    sin(arccos(-6/7))

    Would it be correct that I figure out arccos(-6/7) first and then plug that answer into sin? That's what I've been doing, but it's out of domain since the answer is around 2.6 when I plug the inverse of cos(-6/7) into the calculator.
    \sin(\arccos(x))=\sqrt{1-x^2}

    This can be shown with the identity \cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)=\sqrt{1-\cos^2(x)}
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Moo View Post
    This can be shown with the identity \cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)={ \color{red}\pm}\sqrt{1-\cos^2(x)}
    One will choose either \sin x =+\sqrt{1-\cos^2 x} if the sine is positive (that is to say x\in[0,\pi]) either \sin x =-\sqrt{1-\cos^2 x } if x lies in [\pi,2\pi] where the sine is negative.


    Now, you've to choose between \sin(\arccos x )=+\sqrt{1-x^2} and \sin(\arccos x )=-\sqrt{1-x^2} (What's the range of x\mapsto\arccos x ? )
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