For complex z showing $\cos(z+ \pi) = -\cos(z)$ As the title states, for complex $z$ I want to show $\cos(z+ \pi) = -\cos(z)$.
My first attempt was to change $\cos$ into $(e^{iz} + e^{-iz}) /2$ but then I figured using the identity $\cos(z) = \cos(x)\cosh(y)+i\sin(x)\sinh(y)$ was better since $\cos(x+ \pi)=-\cos(x)$ for real $x$. But now I'm unsure how to proceed.

The sleek, more complex-analytic way to do it: Let $f(z) = \cos z + \cos(z+\pi)$. Then $f$ is entire, and $f(x) = 0$ for $x \in \mathbb{R}$. Hence, by the identity theorem, $f(z) = 0$ for all $z \in \mathbb{C}$.
