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Math Help - Simplify Inverse Trig Function

  1. #1
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    Dec 2008
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    Simplify Inverse Trig Function

    Again, i had mono and got behind...thank you

    Smplify the expression for f(x):

    1. f(x) = sin(cos^-1(x) + sin^-1(x))

    2. f(x) = tan(pi + sin^-1(x))


    sorry, i am not good at typing math...thanks for all your help
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  2. #2
    Super Member
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    Jun 2008
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    1. Easily done using the identities:
    \sin{(A \pm B)} = \sin{A}\cos{B} \pm \sin{B}\cos{A}

    \sin^2{\theta}+\cos^2{\theta} = 1

    ============

    Let \alpha = \arccos{x} \implies \cos{\alpha}=x and let \beta = \arcsin{x} \implies \sin{\beta}=x

    \sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}

    = \sqrt{1-\cos^2{\alpha}}\sqrt{1-\sin^2{\beta}}+\sin{\beta}\cos{\alpha}

    = \sqrt{1-x^2}\sqrt{1-x^2}+x^2

    = 1

    2. Recall that -\tan{x} = \tan{(\pi-x)} and \tan^2{x}+1=\sec^2{x}

    ==============

    \tan{(\pi-(-\arcsin{x}))} = -\tan{(-\arcsin{x})}=\tan{(\arcsin{x})}

    Let \alpha = \arcsin{x} \implies \sin{\alpha} = x

    \tan{\alpha} = \sqrt{\frac{1}{\cos^2{x}}-1} = \sqrt{\frac{1}{1-\sin^2{\alpha}}-1}

    Simply replace and you're done.
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