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Math Help - Inverse Trigonometry

  1. #1
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    Inverse Trigonometry

    Hello all,
    just after a bit of help with this question, do i show this by taking the derivative? i'm not sure..

    Show that
    \tan^{-1}x=\sin^{-1}\frac{x}{\sqrt {1+x^2}} for all x \in (-\infty, \infty)
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  2. #2
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    \tan v = \frac{\sin v}{ \cos v}=\frac{x}{1} \implies \arctan \frac{x}{1} = \arcsin \frac{x}{\sqrt{1+x^2}}

    As you can see above, \sin v = x and \cos v = 1. The hypotenuse then becomes \sqrt{1+x^2} (Pythagoras' theorem).
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  3. #3
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    Hello, Robb!

    No derivatives needed . . . It's an Identity problem.


    Show that: . \tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt {1+x^2}}\right) . for all x \in (\text{-}\infty, \infty)
    \text{Let: }\:\theta \:=\:\tan^{\text{-}1}x .[1]

    Then we have: . \tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}

    \theta is in a right triangle with: opp = x,\;adj = 1

    Using Pythagorus, we have: . hyp = \sqrt{1+x^2}

    Hence: . \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{1+x^2}} \quad\Rightarrow\quad \theta \:=\:\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right) .[2]


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