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Math Help - Inverse trigonometry?

  1. #1
    Super Member fardeen_gen's Avatar
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    Inverse trigonometry?

    Prove that:
    2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]
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  2. #2
    Junior Member Infophile's Avatar
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    May 2009
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    We can differentiate each term and show that f'(x)=g'(x) for all x.

    Thus f(x)=g(x)+C and taking x=0 give C=0.

    There are many calculations but it isn't difficult for all that.

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