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Math Help - 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 -
    \cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)<br />
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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 -
    \cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)
    for the second one rewrite it as cos\bigg(arctan\bigg(\frac{5}{12}\bigg)+arctan\big  g(\frac{12}{5}\bigg)\bigg)..then split it apart using cos(x+y)=cos(x)\cdot{cos(y)}+sin(x)\cdot{sin(y)} and use the fact that 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.

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

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

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

    So, find an expression of \cos y.

    Then, find tan y using the relation: \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
    tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}
    Just a quick note here.

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