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Thread: Real and Imaginary parts

  1. #1
    Member kjchauhan's Avatar
    Nov 2009

    Real and Imaginary parts

    Please help me to solve the following :

    Prove that:

    Sin^{-1}(cosec \alpha) = \left( 2n + (-1)^n \right) \frac{\pi}{2} + i(-1)^n log cot\left(\frac{\alpha}{2} \right)

    Given that,

    Sin^{-1}(a+ib)=n\pi + (-1)^n sin^{-1}(a+ib), n\in Z

    Thanks in advance..

    I got some idea...

    Let sin^{-1}(cosec\alpha)=x+iy

     \therefore cosec\alpha = sin(x+iy)

     \therefore cosec\alpha = sin x cosh y + i sinh y cos x

    on comparing,

    cosec \alpha =sin x cosh y and 0=sinh y cos x

    from these, we have,

    cos x = 0 \Rightarrow x = \frac{\pi}{2}


    cosec^2\alpha = (1-cos^2 x)coshy

    \Rightarrow cosec\alpha = coshy

    \Rightarrow y= cosh^{-1}(cosec\alpha)

    \Rightarrow y= log(cosec\alpha+cot\alpha)

    \Rightarrow y= log(cot\frac{\alpha}{2})

    Hence substituting these values of x and y in above we get the result..
    Last edited by kjchauhan; Jan 7th 2011 at 03:08 PM. Reason: Merged posts.
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