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Thread: inverse trigonometric functions

  1. #1
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    inverse trigonometric functions

    Help needed on these two problems -

    1. Prove that

    tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

    2. Find the exact value of -
    $\displaystyle \cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)
    $
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by darkangel View Post
    Help needed on these two problems -

    1. Prove that

    tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

    2. Find the exact value of -
    $\displaystyle \cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)$
    for the second one rewrite it as $\displaystyle cos\bigg(arctan\bigg(\frac{5}{12}\bigg)+arctan\big g(\frac{12}{5}\bigg)\bigg)$..then split it apart using $\displaystyle cos(x+y)=cos(x)\cdot{cos(y)}+sin(x)\cdot{sin(y)}$ and use the fact that $\displaystyle cos(arctan(x))=\frac{1}{\sqrt{1+x^2}}$ to finish
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  3. #3
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    The second one should be easy.

    $\displaystyle tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$
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  4. #4
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    1. Set $\displaystyle y = \sin^{-1} \frac{x}{\sqrt{1+x^{2}}}$. This implies:

    $\displaystyle \sin y = \frac{x}{\sqrt{1+x^{2}}}$

    Using the identity: $\displaystyle \sin^{2} x + \cos^{2} x = 1 \Rightarrow \cos x = \sqrt{1 - \sin^{2} x}$

    So, find an expression of $\displaystyle \cos y$.

    Then, find tan y using the relation: $\displaystyle \tan y = \frac{\sin y}{\cos y}$ and solve for y.
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  5. #5
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    aite Thanks for everyone's help
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  6. #6
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    Quote Originally Posted by galactus View Post
    $\displaystyle tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$
    Just a quick note here.

    Observe that $\displaystyle \left( {\tan ^{ - 1} x} \right)' + \left( {\cot ^{ - 1} x} \right)' = 0,$ so $\displaystyle \tan ^{ - 1} x - \frac{\pi }
    {4} + \cot ^{ - 1} x - \frac{\pi }
    {4} = 0,$ and the result follows.
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  7. #7
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    woah ! thanks Krizalid !!
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