<s>[INST] Proving that $x=\arccos(\sqrt{\sin\theta})$ is $\sin(x+iy)=\cos\theta+i\sin\theta$ Given:
$$\sin(x+iy)=\cos\theta+i\sin\theta$$
To prove:
$$x=\arccos (\sqrt{\sin\theta})$$
How I tried:
$$\begin{align*}
\sin x \cosh y &= \cos\theta \\
\cos x \sinh y &= \sin\theta
\end{align*}$$
Then tried to use logarithm of hyperbolic complex number.
Also various trignometric form manipulation but I can't get the answer.
 [/INST] given:

$sin(x+iy)=e^{i\theta}$
$sin(x)*cosh(y)+icos(x)*sinh(y)=cos(\theta)+isin(\theta)$
comparing the the real and imaginary part
$\implies sin(x)*cosh(y)=cos(\theta)$ 
&
$\implies cos(x)*sinh(y)=sin(\theta)$
consider hyperbolic form of $sin^2y +cos^2y=1 $
$\cosh^2y-sinh^2y=1$
put values of $sinh(y)  \& cosh(y)$ in above equation
$ \frac {\cos^2\theta}{ \sin^2x} $-$\frac{sin^2\theta}{cos^2x}$=1
cross muliplying and arranging
$cos^2\theta*cos^2x-sin^2\theta*sin^2x=sin^2x*cos^2x$
$(1-sin^2\theta)*cos^2x-sin^2\theta*sin^2x=cos^2x*(1-cos^2x)$
$cos^2x-sin^2\theta(cos^2x+sin^2x)=cos^2x-cos^2x*cos^2x $
$-sin^2\theta=-cos^4x$
taking twice squre root
$cosx=\sqrt\sin\theta$
</s>