Question

Consider complex number $z=\frac{1-isin\theta }{1+icos\theta }$
The value of $sin2\theta$ for argument of 45 degree.

Answer 1

$z=\frac{1-i\sin \theta }{1+i\cos \theta }$

$\Rightarrow argz=arg(1-i\sin \theta )-arg(1-i\cos \theta )$

$\Rightarrow \frac{\pi }{4}={\tan }^{-1}(-\sin \theta )-{\tan }^{-1}\cos \theta$

${\tan }^{-1}\sin \theta +{\tan }^{-1}\cos \theta =\frac{-\pi }{4}$

${\tan }^{-1}\left(\frac{\sin \theta +\cos \theta }{1-\sin \theta \cos \theta }\right)=\frac{-\pi }{4}$

$\frac{\sin \theta +\cos \theta }{1-\sin \theta \cos \theta }=\frac{-\tan \pi }{4}=-1$

$\Rightarrow \sin \theta +\cos \theta +-1\sin \theta \cos \theta$ ( squaring both sides )

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

$\Rightarrow 1+2\sin \theta \cos \theta =1+{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta$

$\Rightarrow 2\sin 2\theta =\frac{4{\sin }^2\theta +{\cos }^2\theta }{4}=\frac{{\sin }^{2}2\theta }{4}$

$\Rightarrow \sin 2\theta =8$

Hence, solved.