For any $z=x+iy$, prove the following:
$$|\sin z| \geq |\sin x|$$
$$|\cos z| \geq |\cos x|$$
$\epsilon$-$\delta $ proof is not required.
I don't really know how to proceed. I know in order to remove the absolute values I can square both sides and I have tried proving this statement using the hyperbolic forms and then the exponential forms but I keep running into circles and I am getting no where…
So for the first one all I have is $$ |\sin z| \geq |\sin x|
$$ $$|\sin z|^2=\sin^2x +\sinh^2y \geq \sin^2x $$ or $${1\over4} |e^{iz}-e^{-iz} |^2≥{1\over4}|e^{ix}-e^{-ix}|^2$$And at this point I think I am beginning to overthink how to square absolute values.
Best Answer
Note that since $\cosh(y)\ge 1$, we have
$$\begin{align}|\sin(z)|^2&=|\sin(x)\cosh(y)+i\cos(x)\sinh(y)|^2\\\\ &=\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y)\\\\ &\ge \sin^2(x) \end{align}$$
and
$$\begin{align}|\cos(z)|^2&=|\cos(x)\cosh(y)+i\sin(x)\sinh(y)|^2\\\\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)\\\\ &\ge \cos^2(x) \end{align}$$