<s>[INST] Proving $|\cos z\ |^2 + |\sin z\ |^2 \geq 1$ My attempt:
For $z :=x+iy$,
$$\cos z =\cos x \cos iy - \sin x \sin iy \\ = \cos x \cosh y - i \sin x \sinh y\\ \sin z = \sin x \cosh y + i \cos x \sinh y$$  
So
$$|\cos z\ |^2 + |\sin z\ |^2 = \cos ^2 x \cosh ^2 y + \sin ^2 x \sinh ^2 y + \cosh^2 y \sin^2x + \cos ^2x \sinh ^2 y\\ = \cos ^2x ( \cosh ^2 y + \sinh ^2 y) + \sin ^2 x (\cosh^2 y + \sinh^2 y) \\ = \cos 2y \leq 1.$$  
I'm not sure where I went wrong, and a numerical check on Wolfram Alpha shows that the inequality should be $\geq$ as suggested.
 [/INST] From $\cosh^2 x\ge1$ and $\sinh^2 x\ge0$, the equations for $\cos z$ and $\sin z$ immediately yield $|\cos z|^2\ge\cos^2x$ and $|\sin z|^2\ge\sin^2x$. The result follows.
</s>