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.

You're almost there:
$$\cos ^2x ( \cosh ^2 y + \sinh ^2 y) + \sin ^2 x (\cosh^2 y + \sinh^2 y)=$$
$$(\cos ^2x + \sin ^2 x) (\cosh^2 y + \sinh^2 y)=$$
$$(\cosh^2 y + \sinh^2 y)=$$
$$2\sinh^2 y + 1\geq1$$
The last step uses the hyperbolic identity $$ \cosh ^2 y - \sinh ^2 y = 1$$ (note the sign difference with the more familiar goniometric one).
