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.

Just continue your approach:
$\cos(z) = \cos(x)\cosh(y)+i\sin(x)\sinh(y)$
Then just plug in $z+\pi$ instead:
$\cos(z+\pi) = \cos(x+\pi)\cosh(y)+i\sin(x+\pi)\sinh(y)$
And observe that both $\cos(x+\pi)=-\cos(x)$ and $\sin(x+\pi)=-\sin(x)$ and you get:
$\cos(z+\pi) = -\cos(x)\cosh(y)-i\sin(x)\sinh(y) = -\cos(z)$
