# How one can prove $\prod_{k=1}^{n-1} (\cos\theta-\cos(k\pi/n)) = 2^{1-n}\sin(n\theta)\csc\theta$?

Can somebody help me to prove this identity?

$$\prod_{k=1}^{n-1} \left(\;\cos\theta-\cos\frac{k\pi}{n}\;\right) = 2^{1-n}\sin(n\theta)\csc\theta$$

I was thinking of using the properties of the complex numbers and the binomial theorem, because I found this problem in the Complex Variables book from Schaum's, but it did not solve my problem.

Does anyone have an idea?

Thanks in advance!

## 2 Answers

Let $x=\sin(n\theta)\csc(\theta)$. I have to show that $x=2^{n-1}\prod_{k=1}^{n-1}(\cos(\theta)-\cos(k\pi/n))$.

Consider $t=e^{i\theta}=\cos\theta+i\sin\theta$. We have $(t-\bar t)\cdot x=(t^n-\bar t^n)$ and hence $$\left(t^n-\frac{1}{t^n}\right)=\left(t-\frac{1}{t}\right)\cdot x.$$ Multiplying $t^n$ on both sides, we then have $t^{2n}-1=(t^2-1)t^{n-1}\cdot x$. On the other hand, $$t^{2n}-1=\prod_{k=0}^{2n-1}(t-\xi^k)=(t^2-1)\prod_{1\le k\le 2n-1,~k\ne n}(t-\xi^k),$$ where $\xi$ is the $2n$-th roots of unity. Note that $\xi=\bar\xi^{-1}$ and $\xi^{n+l}=\bar\xi^{n-l}$ for $1\le l\le n-1$. Thus
$$\prod_{1\le k\le 2n-1,~k\ne n}(t-\xi^k)=\prod_{k=1}^{n-1}(t-\xi^k)(t-\bar\xi^k)=\prod_{k=1}^{n-1}(t^2-(\xi^k+\bar\xi^k)t+1)\\ =t^{n-1}\prod_{k=1}^{n-1}\left(t-(\xi^k+\bar\xi^k)+\frac{1}{t}\right)=t^{n-1}\prod_{k=1}^{n-1}\left(t-(\xi^k+\bar\xi^k)+\bar t\right)\\ =t^{n-1}\cdot 2^{n-1}\prod_{k=1}^{n-1}\left(\cos(\theta)-\cos\left(\frac{k\pi}{n}\right)\right).$$ Hence the assertion $x=2^{n-1}\prod_{k=1}^{n-1}(\cos(\theta)-\cos(k\pi/n))$ follows.

• $$t^{2n}-1=\prod_{k=0}^{2n-1}(t-\xi^k)$$ is sort of obvious, but I had forgotten it and had to look it up. – Keith McClary Jan 7 '18 at 5:10

The Chebyshev polynomials of the second kind satisfy the relation $$U_n(\cos(\theta)) = \frac{\sin((n+1)\theta)}{\sin(\theta)}$$

Hence $U_{n-1}(\cos(\theta)) = \sin(n\theta)\csc(\theta)$.

Note that $\prod\limits_{k=1}^{n-1}(x-\cos(k\pi/n))$ is a polynomial of degree $n-1$ which has $n-1$ distinct zeros corresponding to $x = \cos(k\pi/n), k = 1, \ldots, n-1$.

But $U_{n-1}(\cos(k\pi/n))=\sin(k\pi)\csc(k\pi/n) = 0, k = 1, \ldots, n-1$.

Hence $U_{n-1}(x)$ is a polynomial of degree $n-1$ with the same zeros as the product above.

This means $\prod\limits_{k=1}^{n-1}(x-\cos(k\pi/n)) =CU_{n-1}(x)$ for some constant $C$ which must be determined.

The coefficient of $x^{n-1}$ in $U_{n-1}(x)$ is $2^{n-1}$ but it is $1$ in the product. Therefore $C = 2^{1-n}$.

Therefore $\prod\limits_{k=1}^{n-1}(x-\cos(k\pi/n)) = 2^{1-n}U_{n-1}(x)$ and after substituting $x$ and using the relation stated above, we find $$\prod\limits_{k=1}^{n-1}(\cos(\theta)-\cos(k\pi/n)) = 2^{1-n}U_{n-1}(\cos(\theta)) = 2^{1-n}\sin(n\theta)\csc(\theta)$$