Prove the addition form $cos(z_1+z_2) = cosz_1cosz_2 -sinz_1sinz_2$

Question

Prove the addition form $cos(z_1+z_2) = cosz_1cosz_2 -sinz_1sinz_2$

I was trying this way:

$cos(z_1+z_2) = \frac{e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}}{2} $.

\begin{align}
cos(z_1+z_2) = \frac{e^{-(y_1+y_2)}[cos(x_1+x_2)+isin(x_1+x_2)]+e^{-(y_1+y_2)}[cos(-x_1-x_2)+isin(-x_1-x_2)]}{2}
\end{align}

how $cosx$ is even and $sinx$ odd, I rewrite like this:
\begin{align}
cos(z_1+z_2) = \frac{e^{-(y_1+y_2)}[cos(x_1+x_2)+isin(x_1+x_2)]+e^{-(y_1+y_2)}[cos(x_1+x_2)-isin(x_1+x_2)]}{2}
\end{align}

But it's getting to big and so complicate, there is another way??

Answer 1

Note that\begin{align}\require{cancel}\cos(z_1)\cos(z_2)-\sin(z_1)\sin(z_2)&=\frac{e^{iz_1}+e^{-iz_1}}2\cdot\frac{e^{iz_2}+e^{-iz_2}}2-\frac{e^{iz_1}-e^{-iz_1}}{2i}\cdot\frac{e^{iz_2}-e^{-iz_2}}{2i}\\&=\frac{e^{i(z_1+z_2)}+\cancel{e^{i(z_1-z_2)}}+\bcancel{e^{i(z_2-z_1)}}+e^{-i(z_1+z_2)}}4+\\&{}\quad+\frac{e^{i(z_1+z_2)}-\cancel{e^{i(z_1-z_2)}}-\bcancel{e^{i(z_2-z_1)}}+e^{-i(z_1+z_2)}}4\\&=\frac{e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}}2\\&=\cos(z_1+z_2).\end{align}