# Thread: Real and Imaginary parts

1. ## Real and Imaginary parts

Prove that:

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

Given that,

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

I got some idea...

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

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

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

on comparing,

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

from these, we have,

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

and

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

$\displaystyle \Rightarrow cosec\alpha = coshy$

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

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

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

Hence substituting these values of x and y in above we get the result..