# Prove that $|\sin z| \geq |\sin x|$ and $|\cos z| \geq |\cos x|$

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.

• If you're referring to abs. value of real number then; $$|x+y|^2 = (x+y)^2$$ – Faraad Armwood Mar 15 '16 at 15:20
• Thank you but it was more towards complex numbers like is $$|e^{iz}-e^{-iz}|^2 = (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})?$$ – xxXx Mar 15 '16 at 15:35

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}

Hint:

use addition formulas for $\sin z=\sin (x+iy)$ and the fact that $$\cos (iy)=\cosh y \qquad \sin (iy)=i\sinh y$$

• Yes i'm aware of these identities..as that is what I used to get to the point where I am at now. What I don't understand is how I can prove that a function of y is greater than a function of x – xxXx Mar 15 '16 at 15:30
• It seems that you have not used correctly the addition formula. – Emilio Novati Mar 15 '16 at 15:41
• If you use the correct formula for addition you find the result show in the answer of @Dr.MV – Emilio Novati Mar 15 '16 at 15:44
• Mine is further simplified than @Dr.MV's answer but thank you – xxXx Mar 15 '16 at 15:56