Prove $|Im(z)|\le |\cos (z)|$ for $z\in \mathbb{C}$. Let $z\in \mathbb{C}$ i.e. $z=x+iy$. Show that $|Im(z)|\le |\cos (z)|$.
My hand wavy hint was to consider $\cos (z)=\cos (x+iy)=\cos (x)\cosh (y)+i\sin (x)\sinh(y)$ then do "stuff".
Then I have $|\cos (z)|=|Re(z)+iIm(z)|$ and the result will be obvious.
Thanks in advance.
I am missing something trivial I know.

$$\cos(z) = \cos(x) \cosh(y) + i \sin(x) \sinh(y) $$
Taking the norm squared:
$$ \cos^2(x) \cosh^2(y) + \sin^2(x) \sinh^2(y)$$
We are left with:
$$ \cos^2(x) \left(\frac{1}{2}\cosh(2y) + \frac{1}{2}\right) + \sin^2(x) \left(\frac{1}{2}\cosh(2y) - \frac{1}{2}\right)$$
Simplifying, we get:
$$ \frac{1}{2} \left(\cosh(2y) + \cos(2x) \right)$$
We might as well suppose $\cos{2x} = -1$  Our goal is to show $|$Im$(z)|^2 $ is smaller than this quantity.
That is,
$$ \begin{align} & & y^2  & \leq \frac{1}{2} \cosh{2y} - \frac{1}{2} \\  \iff & & y^2 &\leq (\sinh y)^2  \\  \iff & & y & \leq \sinh y \ \ \ \ \ \ \ \ \forall y\geq0\end{align}$$
A quick computation of the derivative shows that $\frac{d}{dy} y = 1$ but $\frac{d}{dy} \sinh y = \cosh{y}$.  If we want to see that $\cosh y \geq 1$, we can differentiate it again and see that $\sinh y \geq 0$.
