Prove some properties of the complex sine and cosine functions

The following definitions extend the sine and cosine functions to take arguments $z \in \mathbb{C}$ :

$\cos z = \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz} - e^{-iz}}{2i}.$

Prove the following formulas, where $u,v,z$ are complex numbers and $z = x + iy$ .

$\sin (u+v) = \sin u \cos v + \cos u \sin v$ .
$\cos (u+v) = \cos u \cos v - \sin u \sin v$ .
$\sin^2 z + \cos^2 z = 1$ .
$\cos (iy) = \cosh y, \qquad \sin(iy) = i \sinh y$ .
$\cos z = \cos x \cosh y - i \sin x \sinh y$ .
$\sin z = \sin x \cosh y + i \cos x \sinh y$ .

Proof. Using the given definition of the sine of complex numbers we have
$\begin{align*} \sin (u+v) &= \frac{e^{i(u+v)} - e^{-i(u+v)}}{2i} \\[9pt] &= \frac{e^{iu}e^{iv} - e^{-iu}e^{-iv}}{2i} \\[9pt] &= \frac{2 e^{iu}e^{iv} - 2 e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{2e^{iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} - 2e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{e^{iu}(e^{iv} + e^{-iv}) - e^{-iu}(e^{iv} + e^{-iv}) + e^{iv}(e^{iu} + e^{-iu}) - e^{-iv}(e^{iu} + e^{-iv})}{2(2i)} \\[9pt] &= \frac{(e^{iu} - e^{-iu})(e^{iv} + e^{-iv})}{2(2i)} + \frac{(e^{iv} - e^{-iv})(e^{iu} + e^{-iu})}{2(2i)} \\[9pt] &= \sin u \cos v + \sin v \cos u. \qquad \blacksquare \end{align*}$
Proof. Similar to part (a) we compute
$\begin{align*} \cos (u+v) &= \frac{1}{2} \big( e^{i(u+v)} + e^{-i(u+v)} \big) \\[9pt] &= \frac{1}{2} \big( e^{iu} e^{iv} + e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2 e^{iu} e^{iv} + 2 e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2e^{iu}e^{iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + 2 e^{-iu}e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( e^{iu}(e^{iv} + e^{-iv}) + e^{-iu}(e^{iv} + e^{-iv}) + e^{iu}(e^{iv} - e^{-iv}) - e^{-iu}(e^{iv} - e^{-iv}) \big) \\[9pt] &= \frac{1}{4} \big( (e^{iu} + e^{-iu})(e^{iv} + e^{-iv}) \big) + \frac{1}{4} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big) \\[9pt] &= \cos u \cos v - \frac{1}{4i^2} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big)\\[9pt] &= \cos u \cos v - \sin u \sin v. \qquad \blacksquare \end{align*}$
Proof. We compute
$\begin{align*} \sin^2 z + \cos^2 z &= \left( \frac{1}{2i} \right)^2 (e^z - e^{-z})^2 + \left( \frac{1}{2} \right)^2 (e^z + e^{-z})^2 \\[9pt] &= -\frac{1}{4} (e^z - e^{-z})^2 + \frac{1}{4}(e^z + e^{-z})^2 \\[9pt] &= \frac{1}{4} \left( (e^z+e^{-z})^2 - (e^z - e^{-z})^2 \right) \\[9pt] &= \frac{1}{4} \left( (e^z + e^{-z} + e^z - e^{-z})(e^z + e^{-z} - e^z + e^{-z}) \right) \\[9pt] &= \frac{1}{4} \left( (2e^z)(2e^{-z}) \right) \\[9pt] &= 1. \qquad \blacksquare \end{align*}$
Proof. The two computations are as follows,
$\begin{align*} \cos(iy) &= \frac{1}{2} \big( e^{i(iy)} + e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2}(e^{-y} + e^y) \\[9pt] &= \frac{1}{2}(e^y + e^{-y}) \\[9pt] &= \cosh y. \\[9pt] \sin (iy) &= \frac{1}{2i} \big( e^{i(iy)} - e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2i} (e^{-y} - e^y) \\[9pt] &= -\frac{1}{2i} (e^y - e^{-y}) \\[9pt] &= i \sinh y. \qquad \blacksquare \end{align*}$
Proof. We have,
$\cos z = \cos (x+iy) = \cos x \cos (iy) - \sin x \sin (iy) = \cos x \cosh y - i \sin x \sinh y. \qquad \blacksquare$
Proof. We have,
$\sin z = \sin(x+iy) = \sin x \cos (iy) + \cos x \sin (iy) = \sin x \cosh y + i \cos x \sinh y. \qquad \blacksquare$