# 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.

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It is not true. For instance, $n=2$ gives us $2 \cos(t)$ which is clearly not decreasing. – user17762 Mar 17 '12 at 2:28
@SivaramAmbikasaran Sorry, I correct the mistake. – 89085731 Mar 17 '12 at 2:30
You didn't; it was still in the title. – joriki Mar 17 '12 at 2:36
@joriki sorry again. – 89085731 Mar 17 '12 at 2:39
I think there is a problem with your derivative. I do not believe the statement $n \tan (nt) - \tan (nt) < 0$ on $(0, \frac{\pi}{2n})$ is true. – user23784 Mar 17 '12 at 2:44

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 $d'(t)<0$
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Use the convexity of $\tan$ on the interval $[0,nt]$ to obtain an inequality linking $\tan(t)$, $\tan(nt)$ and $\tan(0)$.

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More detail?i need your help – 89085731 Mar 17 '12 at 3:16
What inequality relating $f(a)$, $f(b)$ and $f(c)$ can you write when $f$ is strictly convex and $a<b<c$? – Generic Human Mar 17 '12 at 3:29
yeah, i know your meaning.Thanks. – 89085731 Mar 17 '12 at 3:32

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 $0< nt<\pi/2$, $$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\sin(nt)}{\sin(t)}\le0\tag{4}$$

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