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Math Help - Inverse Trig Identity

  1. #1
    Junior Member Greymalkin's Avatar
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    Inverse Trig Identity

    The actual problem is: prove that sin^-1x=tan^-1{x\over {\sqrt{1-x^2}}

    My problem is: how does sin(tan^-1{x\over \sqrt{1-x^2}}) equal x??
    I understand how sin(sin^-1x)=x but not the rational inverse, why is it so???
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  2. #2
    MHF Contributor
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    Re: Inverse Trig Identity

    The idea is that if x = \sin\alpha, then \sin^{-1}x=\alpha. We have \frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}=\tan\alpha, so \tan^{-1}\frac{x}{\sqrt{1-x^2}} is indeed \alpha.

    One has to take care of various subtleties such as \sqrt{x^2}=|x| rather than x and \sin^{-1}(\sin\alpha)=\alpha only for \alpha\in[-\pi/2,\pi/2].
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  3. #3
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    Re: Inverse Trig Identity

    Hello, Greymalkin!

    \text{Prove: }\:\sin^{\text{-}1}\!x\:=\:\tan^{\text{-}1}\!\left(\frac{x}{\sqrt{1-x^2}}\right)

    \text{Let }\,\theta \:=\:\sin^{\text{-}1}\!x \;\;[1]

    \text{Then: }\,\sin\theta \:=\:x \:=\:\frac{x}{1} \:=\:\frac{opp}{hyp}

    \text{We see that }\theta\text{ is in a right triangle with: }\:\begin{Bmatrix}opp \:=\:x \\ hyp \:=\:1 \end{Bmatrix}
    \text{From Pythagorus: }\:adj \:=\:\sqrt{1-x^2}

    \text{Then: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{x}{\sqrt{1-x^2}}

    \text{Hence: }\:\theta \:=\:\tan^{\text{-}1}\!\left(\frac{x}{\sqrt{1-x^2}}\right) \;\;[2]


    \text{Equating [1] and [2]: }\:\sin^{\text{-}1}\!x \;=\;\tan^{\text{-}1}\!\left(\frac{x}{\sqrt{1-x^2}}\right)
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