Question:
Let $P(x)$ be a polynomial satisfying $$P(k)=\cos{\dfrac{2k\pi}{n}},k=1,2,\cdots,n$$
Show that
$$\deg{P(x)}\ge n-1$$
I want to consider $$P(m+1)-P(m)=\cos{\dfrac{2(m+1)\pi}{n}}-\cos{\dfrac{2m\pi}{n}}=2\sin{\dfrac{\pi}{n}}\sin{\dfrac{2m+1\pi}{n}}$$ Maybe this can use mean value theorem for divided differences Or can we also prove $$P^{(n-1)}(x)\neq 0?$$ But I can't figure out how. Thank you.
Let $\Delta$ be the finite difference operator, $\Delta f(x)=f(x+1)-f(x)$.
Then, as you've shown,
$$\Delta P(m) = -2 \sin \frac{\pi}{n}\sin \frac{(2m+1)\pi}{n}$$
whenever $1\leq m \leq n-1$.
Similarly, we have
$$ \Delta^2P(m)=-2\sin\frac{\pi}{n}\left(\sin\frac{(2m+3)\pi}{n}-\sin\frac{(2m+1)\pi}{n}\right)=-2\sin\frac{\pi}{n}\left(2\sin\frac{\pi}{n}\cos\frac{(2m+2)\pi}{n}\right) $$
for $1 \leq m \leq n-2$;
$$ \Delta^3P(m)=-4\sin^2\frac{\pi}{n}\left(\cos\frac{(2m+4)\pi}{n}-\cos\frac{(2m+2)\pi}{n}\right)=8\sin^3\frac{\pi}{n}\sin\frac{(2m+3)\pi}{n} $$ for $1 \leq m \leq n-3$; and so on.
Continuing in this manner, it is convenient to define $$ \operatorname{trig}_k m=\begin{cases} \cos m && k\text{ even}\\ \sin m && k\text{ odd} \end{cases}$$ Then it follows by induction that $$ \Delta^k(m)=(-1)^{\lceil k/2 \rceil}2^k\sin^k\frac{\pi}{n}\operatorname{trig}_k \frac{(2m+k)\pi}{n} $$ for $1 \leq m \leq n-k$.
In particular,
\begin{eqnarray} \Delta^{n-1}(1)&=&(-1)^{\lceil (n-1)/2 \rceil}2^{n-1}\sin^{n-1}\frac{\pi}{n}\operatorname{trig}_{n-1}\frac{(n+1)\pi}{n}\\ &=&(-1)^{\lceil (n+1)/2 \rceil}2^{n-1}\sin^{n-1}\frac{\pi}{n}\operatorname{trig}_{n-1}\frac{\pi}{n} \end{eqnarray} can easily be seen to be nonzero for $n>1$: $\sin \frac{\pi}{n}$ only vanishes when $n=1$, $\cos \frac{\pi}{n}$ only vanishes when $n=2$, and when $n=2$ the formula involves $\operatorname{trig}_1 \frac{\pi}{2}=\sin \frac{\pi}{2}$.
If $n=1$, the problem statement is trivially true. But if $n>1$, we have just shown that the $(n-1)^{\rm st}$ finite difference of $P$ doesn't vanish everywhere. It follows that, if $P$ is a polynomial, it must have degree at least $n-1$.
