Question

Seperate ${\sin }^{-1}(\cos \theta +i\sin \theta )$ into real and imaginary parts, where $\theta$ is a positive acute angle.

Answer 1

$\Rightarrow \cos \theta +i\sin \theta =\sin (x+iy)$
We know that
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
$\Rightarrow \cos \theta +i\sin \theta =\sin x\cos iy+\cos x\sin iy$
We know that $\cos iy=\cos hy$ and $\sin iy=i\sin hy$
$\Rightarrow \cos \theta +i\sin \theta =\sin x\cos hy+i\cos x\sin hy$
Equating the real and imaginary parts, we get
Real part $=\cos \theta =\sin x\cos hy$ ..............$\left(1\right)$
Imaginary part $=\sin \theta =\cos x\sin hy$ ..............$\left(2\right)$
Squaring and adding $\left(1\right)$ and $\left(2\right)$ we get
${\cos }^2\theta +{\sin }^2\theta ={\sin }^{2}x{\cosh }^{2}y+{\cos }^{2}x{\sinh }^{2}y$
$\Rightarrow 1={\sin }^{2}x{\cosh }^{2}y+{\cos }^{2}x{\sinh }^{2}y$
Take ${\cosh }^{2}y=1+{\sinh }^{2}y$
$\Rightarrow 1={\sin }^{2}x(1+{\sinh }^{2}y)+{\cos }^{2}x{\sinh }^{2}y$
$\Rightarrow 1={\sin }^{2}x+{\sin }^{2}x{\sinh }^{2}y+{\cos }^{2}x{\sinh }^{2}y$
Taking common, we get
$\Rightarrow 1={\sin }^{2}x+{\sinh }^{2}y({\sin }^{2}x+{\cos }^{2}x)$
$\Rightarrow 1={\sin }^{2}x+{\sinh }^{2}y$ $\because {\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow 1-{\sin }^{2}x={\sinh }^{2}y$
$\Rightarrow {\cos }^{2}x={\sinh }^{2}y$
From $\left(2\right)$ we have
$\sin \theta =\cos x\sin hy$
or $\sin hy=\frac{\sin \theta }{\cos x}$
$\Rightarrow {\cos }^{2}x={\left(\frac{\sin \theta }{\cos x}\right)}^2$$\Rightarrow {\cos }^{4}x={\sin }^2\theta$
or ${\cos }^{2}x=\sin \theta$
$\because \theta$ being a positive acute angle, $\sin \theta$ is positive.
As $x$ is to be between $-\frac{\pi }{2}$ and $\frac{\pi }{2},\therefore$ we have$\cos x=\sqrt{\sin \theta }$ or $x={\cos }^{-1}\left(\sqrt{\sin \theta }\right)$
The relation then eqn$\left(2\right)$ gives
$\sin hy=\sqrt{\sin \theta }$ so that $y=\log [\sqrt{\sin \theta }+\sqrt{(1+\sin \theta )}]$