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.
Best Answer
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 $k\ge1$.