# If $z=x+iy$, then $|\cos z|\ge |\cos x|$ and $|\sin z|\ge \sin x$

I'm trying to use that

$$|\cos z| = \left|\frac{1}{2}(e^{iz}+e^{-iz})\right| = \left|\frac{1}{2}(\cos z+i\sin z + \cos -z + i\sin -z)\right| = \left|\frac{1}{2}(2\cos z)\right|$$

but I return to the same thing. Also, I couldn't understand this related question's answers, specially in why

$$\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y)\\ \ge \sin^2(x)$$

• Because $\cosh^2-1 = \sinh^2$ – user384138 Nov 3 '16 at 16:02

## 1 Answer

Hint. One may write\begin{align} |\cos(z)|^2&=|\cos(x)\cosh(y)+i\sin(x)\sinh(y)|^2\\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\\ &=\cos^2(x)\cosh^2(y)+(1-\cos^2(x))\sinh^2(y)\\ &=\cos^2(x)(\cosh^2(y)-\sinh^2(y))+\sinh^2(y)\\ &=\cos^2(x)\cdot 1+\sinh^2(y)\\ &\ge \cos^2(x), \end{align} and similarly for $|\sin(z)|^2$.