Proving that $0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)$ I need help proving that
$$
0 \leq k \sin\left(\frac{2\pi}{n}\right) - \sin\left(\frac{2\pi k}{n}\right)
$$
For $n > 2$ and positive $k$.
I've tried all sorts of identities, and nothing have worked. Any help would be greatly appreciated.
The things I have tried are the following:
Using a identity for multiples angle formula:
$$
k\sin \left(\frac{2 \pi}{n}\right)  >\sin\left(k\cdot\frac{2 \pi }{n}\right)=2^{k -1}\prod_{i=0}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
$$
k \geq 2^{k -1}\prod_{i=1}^{k-1}\sin \left(\frac{\pi i}{k}+\frac{2 \pi }{n}\right)
$$
Which I could not make sense of.
I also tried rewriting the inequality
$$
k\sin \left(\frac{2 \pi}{n}\right)  +\sin \left(\frac{2 \pi (n -k)}{n}\right)>0
$$
Which got me here:
$$\begin{split}
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi (n -k)}{n}\right)\\
&=(k-1)\sin \left(\frac{2 \pi}{n}\right)+2\sin \left(\frac12\left[\frac{2 \pi}{n}+\frac{2 \pi (n -k)}{n}\right]\right)\sin \left(\frac12\left[\frac{2 \pi}{n}-\frac{2 \pi (n -k)}{n}\right]\right)
\end{split}$$
But I could not make that work either.
 A: I am assuming that $k,n\in\mathbb{Z}$.
Suppose that for some $k$, we have
$$
\sin\left(\frac{2\pi k}{n}\right)\le k\sin\left(\frac{2\pi}{n}\right)\tag1
$$
Note that $(1)$ is trivially true for $k=1$.
For $k\ge2$, we have
$$
k\,\overbrace{\sin\left(\frac{2\pi}4\right)}^1\gt k\,\overbrace{\sin\left(\frac{2\pi}3\right)}^{\sqrt3/2}\gt1\ge\overbrace{\sin\left(\frac{2\pi k}{n}\right)}^{\le1}\tag2
$$
Thus, $(1)$ is true for $n=3$ and $n=4$. Therefore, assume, $n\ge5$.
$$
\begin{align}
\sin\left(\frac{2\pi(k+1)}{n}\right)
&=\sin\left(\frac{2\pi k}{n}\right)\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\sin\left(\frac{2\pi}{n}\right)\tag{3a}\\
&\le k\sin\left(\frac{2\pi}{n}\right)\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\sin\left(\frac{2\pi}{n}\right)\tag{3b}\\
&=\left(k\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{2\pi k}{n}\right)\right)\sin\left(\frac{2\pi}{n}\right)\tag{3c}\\
&\le(k+1)\sin\left(\frac{2\pi}{n}\right)\tag{3d}
\end{align}
$$
$\text{(3a):}$ sine of a sum identity
$\text{(3b):}$ apply $(1)$ with $\cos\left(\frac{2\pi}n\right)\ge\cos\left(\frac{2\pi}5\right)=\frac{-1+\sqrt5}4$
$\text{(3c):}$ distribute a product over a sum
$\text{(3d):}$ $\cos(x)\le1$ and $\sin\left(\frac{2\pi}n\right)\ge0$ for $n\ge2$
Inequality $(1)$ is true for $k=1$. Then $(3)$ and induction prove $(1)$ for any $k\ge1$.
A: Assume $0 < k \le 1$ and $n \ge 4$. Put $x = \dfrac{2\pi k}{n} \implies k = \dfrac{nx}{2\pi}$. Thus you prove: $\dfrac{\sin\left(\frac{2\pi}{n}\right)}{\frac{2\pi}{n}} \le \dfrac{\sin x}{x}$. But the function $ f(x) = \dfrac{\sin x}{x}$ on $\left(0,\frac{2\pi}{n}\right]$ has $f’(x) = \dfrac{x\cdot \cos x - \sin x}{x^2} < 0$ since $\tan x > x$ on this same domain of $x$. Thus $f$ is a decreasing function and you have $f(x) \ge f\left(\frac{2\pi}{n}\right)$ since $0 < x < \frac{2\pi}{n}$. Done.
Note:The inequality reverses when $k < 1$.
A: It is a consequence of convexity of $\sin$ on $[0,\pi]$*. The straight line from $(0,0)$ to $(x,\sin(x))$ passes through the point $(x/k,\sin(x)/k)$. Therefore, we have $\sin(x)/k\le \sin(x/k)$. Now set $x=\frac{2\pi k}{n}$.
*We must have $x\in [0,\pi]$ to use convexity, so this unfortunately only gives the result for $n\ge2k$.
