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.

One of my favourite trig. identities is
\begin{align}\lvert \sin{(x+iy)} \rvert^2 &= \lvert \sin{x}\cosh{y} + i\cos{x}\sinh{y} \rvert^2 \\
&= \sin^2{x}\cosh^2{y}+\cos^2{x}\sinh^2{y} \\
&= \sin^2{x}(1+\sinh^2{y})+(1-\sin^2{x})\sinh^2{y} \\
&= \sin^2{x}+\sinh^2{y}.\end{align}
To get the equivalent for $\lvert \sin{(x+iy)} \rvert^2$, replace $x$ by $\pi/2-x$ and $y$ by $-y$, which just swaps the sines and cosines in the proof (and the sign of the imaginary term, which is irrelevant). So
$$ \lvert \cos{(x+iy)} \rvert^2 =  \cos^2{x}+\sinh^2{y}, $$
and then the sum is
$$ \lvert \sin{(x+iy)} \rvert^2 + \lvert \cos{(x+iy)} \rvert^2 =  1+2\sinh^2{y} \geq 1 $$
since the other term is the square of a real.
