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Math Help - inverse trig function

  1. #1
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    inverse trig function

    simplify: \tan\left(2\cos^{-1}(\frac{x}{4})\right)

    using the double angle foruma  tan(2x) = \frac{2tan(x)}{1-tan^2x}
    i got to this part:

    \frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    simplify: \tan\left(2\cos^{-1}(\frac{x}{4})\right)

    using the double angle foruma  tan(2x) = \frac{2tan(x)}{1-tan^2x}
    i got to this part:

    \frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}
    Let \theta = \cos^{-1} \left( \frac {x}{4} \right)

    \Rightarrow \cos \theta = \frac {x}{4} (See diagram below)


    Then \tan \left( 2 \cos^{-1} \left( \frac {x}{4} \right) \right) = \tan (2 \theta )

    = \frac {2 \tan \theta}{1 - \tan^2 \theta}

    Can you find the value of  \tan \theta from the diagram below? ......Of course you can! The problem is child's play from here. (You didn't have to use a triangle, you could have used the Pythagorean Identities for sine and cosine)
    Attached Thumbnails Attached Thumbnails inverse trig function-tan.gif  
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