The real part of $e^{e^{i\theta}}$?

Question

I know that a solution for this question exists on SE, I have seen it, but it doesn’t satisfy my query. I couldn’t understand the solution.

Here is an excerpt from one of the answers

$$z = e^{e^{i \theta} }$$

$$\implies z=e^{\cos \theta +i\sin \theta }$$

$$\implies z=e^{\cos \theta} e^{i\sin \theta }$$

$$\implies z=e^{\cos \theta}(\cos (\sin\theta) +i\sin(\sin \theta))$$

I didn’t understand the last line of the answer. Up to the third line, everything is clear.

Answer 1

By Euler's identity we have

$$ e^{i\sin \theta }=\cos (\sin \theta)+i\sin(\sin\theta)$$

Answer 2

$e^{it}=\cos t+i\sin t$ for any real number $t$. Put $t=\sin \theta$ and you will get the last step.

Answer 3

You can extend the Euler identity to any function: $$ e^{if(x)} = \cos(f(x))+i\sin(f(x)) $$ Thus, for $ f(x) = \sin(x) $ you will have the obtained value. The real part is then: $$ \mathrm{Re}\{z\} = \mathrm{Re}\{e^{\cos\theta}(\cos(\sin(\theta))+i\sin(\sin(\theta)))\} = \mathrm{Re}\{e^{\cos\theta}\cos(\sin(\theta))+ie^{\cos\theta}\sin(\sin(\theta))\} = $$ $$ = e^{\cos\theta}\cos(\sin(\theta)) $$ I hope the answer is satisfactory.