Sum and product trigonometry

Using de Moivre's Theorem, we see that $\sin(2n+1)x \; = \; \sum_{j=0}^n (-1)^j \binom{2n+1}{2j+1}\cos^{2n-2j}x\, \sin^{2j+1}x \; = \; \sin x f\big(\sin^2x\big)$ where $f(X)$ is the degree $n$ polynomial $f(X) \; = \; \sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}(1-X)^{n-j}X^j$ Note that the zeros of $f(X)$ are thus $\sin^2\tfrac{k\pi}{2n+1}$ for $k = 1,2,...,n$ .

The coefficient of $X^n$ in $f(X)$ is $(-1)^n \sum_{j=0}^n \binom{2n+1}{2j+1} \; = \; (-1)^n 2^{2n}$ while the coefficient of $X^{n-1}$ is $\begin{aligned} \sum_{j=0}^{n-1}(-1)^j \binom{2n+1}{2j+1} (-1)^{n-j-1}(n-j) & = \; \tfrac12(-1)^{n-1}\sum_{j=0}^{n-1}\frac{(2n+1)!}{(2j+1)!(2n-2j-1)!} \; = \; \tfrac12(-1)^{n-1}(2n+1)\sum_{j=0}^{n-1}\binom{2n}{2j+1} \\ & = \; \tfrac12(-1)^{n-1}(2n+1)2^{2n-1} \; = \; (-1)^{n-1}(2n+1)2^{2n-2} \end{aligned}$ Thus we deduce that $\sum_{k=1}^n \left(2\sin\tfrac{k\pi}{2n+1}\right)^2 \; = \; 4\frac{(2n+1)2^{2n-2}}{2^{2n}} \; = \; 2n+1$ On the other hand, the coefficient of $X_0$ in $f(X)$ is $\binom{2n+1}{1} = 2n+1$ , and hence $\prod_{k=1}^n \left(2\sin\tfrac{k\pi}{2n+1}\right)^2 \; = \; 2^{2n} \frac{2n+1}{2^{2n}} \; = \; 2n+1$ Thus both expressions are equal to $2n+1$ .

@Pi Han Goh

Proving the equation is identical to proving that $4 \sum_{k=1}^n \sin^2 \left( \dfrac {k \pi}{2n+1} \right) = 4^n \prod_{k=1}^n \sin^2 \left (\dfrac{k \pi}{2n+1} \right) .$

Using the double angle formula, $\sin^2 (A) = \frac12 (1 - \cos(2A) )$ , the left hand side can be expressed as

$4 \sum_{k=1}^n \left [ \dfrac12 \left (1 - \cos\left( \dfrac {2k \pi}{2n+1} \right) \right) \right ] = 2 \left [ n - \underbrace{\sum_{k=1}^n \cos\left( \dfrac {2k \pi}{2n+1} \right)}_{=-1/2} \right ] = 2n+1.$

The final summation above follows from the trigonometric identity:

$1 + 2\sum_{k=1}^n \cos (k\theta) = \sum_{k=-n}^n e^{k i \theta} = \dfrac{e^{(n+ 1/2)\theta} - e^{-i(n-1/2)\theta} }{e^{i\theta/2} - e^{-i \theta /2}} = \dfrac{\sin[(n+\frac12)\theta]}{\sin(\frac\theta2)}$

In this case, $\theta = \frac{2\pi}{2n+1}$ , so this summation simplifies to $-\frac12$ .

On the other hand, using the complementary formula, $\sin(A) = \sin(\pi - A)$ , the right hand side can be expressed as

$4^n \prod_{k=1}^{2n} \sin \left (\dfrac{k \pi}{2n+1} \right).$

This product of sines is also another famous trigonometric identity, which simplifies to $\frac{2n+1}{2^{2n}}$ . This can proven using the identity $\displaystyle \prod_{k=1}^{n-1}\sin\frac{k\pi}n = \dfrac n{2^{n-1}}$ , one of its many proofs can be found here .

Thus, the right hand side simplifies to

$4^n \cdot \dfrac{2n+1}{2^{2n}} = 2n+ 1$

as well. Hence, the trigonometric identity in question is true.

Yes, both expressions are equal to $2n+1$ . It seems that the easiest proof is to notice that $(2\sin(k\pi/n))^2$ are the non-zero eigenvalues with multiplicity 2 of the circulant matrix $\begin{pmatrix}2 & -1 & 0 & \dots & -1 \\ -1 & 2 & -1 & 0 & \dots \\ 0 & -1 & 2 & -1 & \dots \\ \dots & \dots & \dots & \dots & \dots \\ -1 & \dots & 0 & -1 & 2\end{pmatrix}.$ Therefore, the sum is half the trace $2n+1$ and the product is the square root of $(2n+1)\times$ the determinant of any principal $2n\times 2n$ minor, which is $2n+1$ too (the determinants satisfy simple three-term recurrence, which alows the calculation). @Pi Han Goh @Mark Hennings