Proving that $x=\arccos(\sqrt{\sin\theta})$ is $\sin(x+iy)=\cos\theta+i\sin\theta$

Question

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.

Answer 1

We start off with two real numbers $x$ and $y$, and consider the complex function $\sin(x+iy)$. As has been pointed out by others in this thread, this is a complex number that can written as $z = \sin(x + iy) = \sin(x)\cosh(y)+i\cos(x)\sinh(y)$.

We are then asked to express this result in terms of a new variable $\theta$ by means of the relation $z = e^{i\theta} = \cos(\theta) + i\sin(\theta)$. In general this is only possible if $\theta$ itself is a complex number, and equally so for the functions in which it appears. For example $\sin(\theta)$ can be written as $\sin(\theta) = \frac{(z - 1/z)}{2i}$.

However, the OP has now stated that $\theta$ is a real. This implies that $z$ lies on the unit circle. Using the fact that the norm of $z$ is equal to one, we see $x$ and $y$ are related by: $\sinh(y) = \pm\cos(x)$. Note that this equation can only have a solution when $\sinh(y)$ is in the interval $(-1, 1)$, hence values of $y$ are restricted to the interval $(-0.881377, +0.881377)$.

The two branches of the solution are described by the following equations.

A) $y = +arc\sinh(\cos(x))$. Also $\sin(\theta) = \cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$

B) $y = -arc\sinh(\cos(x))$. Also $\sin(\theta) = -\cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$

We see that in both branches $y$ and $\theta$ are periodic functions of $x$. If we combine the two branches in a single plot, $y$ and $\theta$ oscillate perfectly in phase, each having period $\pi$. The amplitude of $y$ is $arc\sinh(1) = 0.881377$ and the amplitude of $\theta$ is $\pi/2$. Note however that this interpretation does not apply to the separate branches, because then the sign of $\theta$ remains the same.

Answer 2

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

$cosx=\sqrt\sin\theta$