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Math Help - inverse of trigonometry

  1. #1
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    inverse of trigonometry

    How did 15 degree come in the following equation.


    http://dc474.4shared.com/download/8_...55515-f1b2071a
    plz help.

    thanks in advance
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  2. #2
    Grand Panjandrum
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    Re: inverse of trigonometry

    Quote Originally Posted by SiddS View Post
    How did 15 degree come in the following equation.


    Untitled.png - 4shared.com - photo sharing - download image
    plz help.

    thanks in advance
    1. It is not 15 but -15 degrees.

    2. Because this involves \sqrt{3} you need to think in terms of 30 or 60 degrees, so look up the tangent half angle formula.

    CB
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  3. #3
    Member sbhatnagar's Avatar
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    Re: inverse of trigonometry

    \theta  = \tan^{-1}{\left( \frac{1-\sqrt{3}}{1+\sqrt{3}}\right)}=\tan^{-1}{\left( \frac{1-\sqrt{3}}{1+(1)(\sqrt{3})}\right)}

    Using the formula: \tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}{\left( \frac{x-y}{1+xy}\right)}

    we get \theta= \tan^{-1}(1)-\tan^{-1}(\sqrt{3})= 45^{\circ}-60^{\circ}=-15^{\circ}
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