Prove for $\cos (x+iy)$ I know it is such a foolish thing to do to ask this dumb question in this site. 
Please prove that $\cos (x + iy) = \cos x \cosh y - i\sin x \sinh y$ and 
$\cos (x - iy) = \cos x \cosh y + i\sin x \sinh y$
I admit that iI don’t even know how to start proving this. Honestly, I don't know much about complex numbers and it’s hard for me to understand the answers on the net.
Please kindly prove this with matching explanations even it will cost down-voting this question. Thank you guys for understanding and patience.
 A: Assuming that you define the cosine on the complex plane by
$$
\cos z=\frac{\exp(iz)+\exp(-iz)}{2}
$$
you have that $\cos$ is a holomorphic function, which coincides with the cosine defined on the real numbers. Therefore, by the uniqueness theorem about holomorphic functions that coincide on a set with an accumulation point, the identity
$$
\cos(z_1+z_2)=\cos z_1\cos z_2-\sin z_1\sin z_2
$$
holds for all $z_1,z_2\in\mathbb{C}$. Of course the same that we said for the cosine can be said for the sine, defined by
$$
\sin z=\frac{\exp(iz)-\exp(-iz)}{2i}
$$
Now, if $y$ is real,
$$
\cos(iy)=\frac{\exp(i^2y)+\exp(-i^2y)}{2}=\frac{\exp y+\exp(-y)}{2}=\cosh y
$$
and
$$
\sin(iy)=\frac{\exp(i^2y)-\exp(-i^2y)}{2i}=-\frac{\exp y-\exp(-y)}{2i}=i\sinh y
$$
and so
$$
\cos(x+iy)=\cos x\cos(iy)-\sin x\sin(iy)=\cos x\cosh y-i\sin x\sinh y.
$$
A: By the definition: $$\begin{align} \cos(x+iy)&=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2}\\
&=\frac{e^{-y+xi}+e^{y-xi}}{2}\\
&=\frac{e^{-y}(\cos(x)+i\sin(x))+e^{y}(\cos(x)-i\sin(x))}{2}\\
&=\frac{\cos(x)(e^{-y}+e^{y})}{2}\ldots\end{align}$$ and the sequal is trivial.
A: $$\cos ix +i\sin ix=e^{-x}$$
also $$\cos ix+i\sin (-ix)=e^{x}$$
adding both $$ 2\cos ix=e^{x}+e^{-x}$$
ie $$\cos ix=\cosh (x)$$ similarly subtracting the two equation gives 
$$i\sin ix=-\sinh (x)$$
in your question use $$\cos (a+b)=\cos a\cos b-\sin a\sin b$$
further substitute above transformations.
