# Supremum and infimum of $\frac{x}{\sin x}$ on $\left[0,\frac{\pi}{2}\right]$

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$\frac{x}{\sin x}$ is positive, increasing and log-convex on $\left(0,\frac{\pi}{2}\right)$. It follows that the minimum of the given function is attained at $x=0$ (where I guess $\frac{x}{\sin x}$ is defined as $1$, removing the removable discontinuity) and the maximum is attained at $x=\frac{\pi}{2}$.

Proof. By letting $g(x)=\log(x)-\log(\sin x)$ we have $g'(x)=\frac{1}{x}-\frac{1}{\tan x}\geq 0$ (since $\tan x\geq x$) and $g''(x)=\frac{1}{\sin^2 x}-\frac{1}{x^2}\geq 0$ (since $\sin x \leq x$).

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### Telaprolu prasanth

Updated on November 12, 2020

What is the supremum and infimum of $\frac{x}{\sin(x)}$ on the interval $[0,\frac{ \pi}{2}]$?
Hint: $\frac{x}{\sin(x)}$ is increasing on the given interval.
Do you mean the interval $(0,\pi/2]$? The expression is undefined at $x=0$