Prove $\frac{\sin(nt)}{\sin(t)}$ is decreasing in $(0,\frac{\pi}{2n})$ I want to prove $\frac{\sin(nt)}{\sin(t)}$ is decreasing in $(0,\frac{\pi}{2n})$, when I differentiate once, I get $n\tan(t)-\tan(nt)$, I want to prove $n\tan(t)-\tan(nt)<0$,but I can't continue.
 A: Use the convexity of $\tan$ on the interval $[0,nt]$ to obtain an inequality linking $\tan(t)$, $\tan(nt)$ and $\tan(0)$.
A: Since $\frac{\sin nt}{\sin t}>0$ on $(0,\frac{\pi}{2n})$, it is decreasing iff $l(t)=\log\frac{\sin nt}{\sin t}$ is decreasing. Let us differentiate it: $d(t)=\frac{d}{dt}l(t)=n\cot nt-\cot t$. We want to show that $d(t)<0$ on $(0,\frac{\pi}{2n})$.


*

*$d(t)\to 0$ as $t\to 0$ (using $\frac{\sin x}{x}\to 1$ or whatever else you want)

*$d'(t)=\frac{1}{sin^2t}-\frac{n^2}{\sin^2nt}$, and, since $\sin$ is concave on $(0, \pi/2)$, $\sin t>\sin nt/n$ for $t\in(0,\pi/2n)$, and $d'(t)<0$

A: First, as you noted,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\frac{\sin(nt)}{\sin(t)}
&=\frac{n\cos(nt)\sin(t)-\sin(nt)\cos(t)}{\sin^2(t)}\\
&=(n\tan(t)-\tan(nt))\frac{\cos(nt)\cos(t)}{\sin^2(t)}\tag{1}
\end{align}
$$
Next use that
$$
\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\ge\tan(x)+\tan(y)\tag{2}
$$
for $\tan(x),\tan(y),\tan(x+y)\ge0$ to deduce that for $0< nt<\pi/2$,
$$
\tan(nt)\ge n\tan(t)\tag{3}
$$
Thus, $(1)$ and $(3)$ imply that, for $0< nt<\pi/2$,
$$
\frac{\mathrm{d}}{\mathrm{d}t}\frac{\sin(nt)}{\sin(t)}\le0\tag{4}
$$
