# Estimate trigonometric functions with complex argument

I would like to prove the following estimates

$\vert \sin(z)\vert\leq \sinh(s)$ and $\vert \cos(z)\vert\leq \cosh ( s )$ ,where $z\in D_s(0)\subset\mathbb{C}$ and $D_s(0)$ denotes the disc with radius $s\geq 0$.

I showed that for $z=r\cdot e^{i\theta}$ $\vert \sin(r\cdot e^{i\theta} ) \vert = \sqrt{sin^2(r\cos(\theta))+\sinh^2( r\sin( \theta ))}$

and similarly

$\vert \cos(r\cdot e^{i\theta} ) \vert = \sqrt{sin^2(r\cos(\theta))+\cosh^2( r\sin( \theta ))}$.

edit: Dr. MV has proven the estimates for $\cos(z), \sin(z)$ below.

Best wishes

• You've used r twice in the middle so change to $z=r e^{i \theta} \in D_s (0)$ with $s \geq 0$. – AHusain Apr 22 '16 at 20:49
• If you know the maximum modulus principle, then I think these follow quite directly. So it may be a good idea to clarify whether you have access to this. – Semiclassical Apr 22 '16 at 21:29
• Thank you, I didn't know that principle. – Hasti Musti Apr 23 '16 at 7:57

Let $f(\theta)$ be given by

$$f(\theta)=\sin^2(r\cos(\theta))+\sinh^2(r\sin(\theta))$$

Then, the derivative $f'(\theta)$ is

$$f'(\theta)=-r\left(\sin(2r\cos(\theta))\sin(\theta)-\sinh(2r\sin(\theta))\cos(\theta)\right) \tag 1$$

Note that $f'(\theta)$ is $\pi$-periodic with $f'(\theta)=0$ for $\theta =n\pi/2$.

Since $|\sin(x)|\le |x|$ and $|\sinh(x)|\ge |x|$, $f'(\theta)$ cannot be zero except at the zeroes of the sine and cosine functions. The maxima of $f(\theta)$ occur, therefore, at the zeroes of the cosine function where $f(\theta)=\sinh^2(r)$.

Finally, inasmuch as the hyperboic sine function is monotonically increasing, $\sinh(r)\le \sinh(s)$ for $s\ge r\ge 0$.

We conclude that for $s\ge r\ge 0$,

$$|\sin(z)|\le |\sinh(s)|$$

And we are done!

• This only appears to prove it for real $x$. But the setting is $\mathbb{C}$. – Semiclassical Apr 23 '16 at 3:13
• @semiclassical Please have a closer look. The inequality is in terms of magnitudes, which are real. $r=|z|$ and $\theta =\arg(z)$. All should be correct. ;-)) – Mark Viola Apr 23 '16 at 3:19
• I see your point. A suggestion: since $|\sin z|=|\sin\overline{z}|=|\sin(-z)|$ one can assume $\text{Re }z,\text{Im }z\geq 0$ (and therefore $\theta\in [0,\pi/2]$) without loss of generality. Hence the use of integer $n$ is unnecessary. – Semiclassical Apr 23 '16 at 3:42
• Thank you Dr. MV! Your proof is very awesome. I worked out your argumentation for $\vert \cos(z)\vert\leq \vert \cosh(r) \vert$ as well. It works fine... :-) – Hasti Musti Apr 23 '16 at 8:02
• Hasti, you're welcome. My pleasure. Pleased to hear this was helpful! -Mark – Mark Viola Apr 23 '16 at 13:55