# If $|z| \leq \pi/2$ and $|\sin z| \leq 1/4$, then $|z| \leq (4 \sin(1/4))^{-1} |\sin z|$

I came across the following assertion and am having trouble justifying it:

If $z$ is a nonzero complex number with $|z| \leq \pi/2$ and $|\sin z| \leq 1/4$, then $$\left| \frac{z}{\sin z} \right| \leq \frac{1/4}{\sin(1/4)} = 1.0104931\ldots$$

I would appreciate some help. Thanks.

EDIT: Andrew has pointed out that the above inequality fails if $z = \sin^{-1}(1/4) = 0.25268025\ldots$. After more thought, I figured out that if $z$ is real with $|z| \leq \pi/2$ and $|\sin z| \leq 1/4$, then $|z| \leq \sin^{-1}(1/4)$ and $z / \sin z$ is increasing in $[0,\sin^{-1}(1/4)]$; consequently, $$\left| \frac{z}{\sin z} \right| \leq \frac{\sin^{-1}(1/4)}{1/4} = 1.0107210\ldots$$ holds.

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Is $z$ complex? –  Américo Tavares Jul 6 '11 at 19:32
The tag says complex numbers so most likely... –  mathmath8128 Jul 6 '11 at 20:04
After all $z$ is real! –  Américo Tavares Jul 7 '11 at 8:55
If $z$ is real why the "complex-numbers"-tag? –  Américo Tavares Jul 7 '11 at 9:10
The last inequality is valid and for complex $z$. –  Andrew Jul 7 '11 at 11:04

It is not true. Function $f(x)=\frac x{\sin x}$ is increasing on $(0,\pi/2)$, so function $$g(x)=\frac{f(x)}{f(\sin x)} =\frac{x \sin (\sin (x))}{ \sin^2(x)}$$ is increasing too since $0<\sin x<x$ and $f(0)=1$. So $g(x)\ge1$ on $[0,\pi/2)$. Taking $z=\sin^{-1}(1/4)$ we have $$g(z)=\frac{\sin^{-1}(1/4)}{1/4}\frac{\sin(1/4)}{1/4}>1.$$