Establishing $\frac{ \sin mx}{\sin x}=(-4)^{(m-1)/2}\prod_{1\leq j\leq(m-1)/2}\left(\sin^2x-\sin^2\frac{2\pi j}{m}\right) $ for odd $m$

Question

Let $m$ be an odd positive integer. Prove that

$$ \dfrac{ \sin (mx) }{\sin x } = (-4)^{\frac{m-1}{2}} \prod_{1 \leq j \leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi j }{m } \right) \right) $$

Atempt to the proof

My idea is to use induction on $m$. The base case is $m=3$ and we obtain

$$ \dfrac{ \sin (3x) }{\sin x } = (-4) ( \sin^2 x - \sin^2 (2 \pi /3 ) ) $$

and this holds if one uses the well known $\sin (3x) = 3 \sin x - 4 \sin^3 x $ identity.

Now, if we assume the result is true for $m = 2k-1$, then we prove it holds for $m=2k+1$. We have

$$ \dfrac{ \sin (2k + 1) x }{\sin x } = \dfrac{ \sin [(2k-1 + 2 )x] }{\sin x } = \dfrac{ \sin[(2k-1)x ] \cos (2x) }{\sin x } + \dfrac{ \cos [(2k-1) x ] \sin 2x }{\sin x } $$

And this is equivalent to

$$ cos(2x) \cdot (-4)^{k-1} \prod_{1 \leq j \leq k-1 }\left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi j }{m } \right) \right) + 2 \cos [(2k-1) x ] \cos x $$

Here I dont see any way to simplify it further. Am I on the right track?

Answer 1

Note that $\sin (x-\frac{2\pi j }{m})=-\sin(x+\frac{(m-2j)\pi}{m})$ and $m-2j$ goes through the odd numbers $1,...m-2$ when $ 1\le j \le \frac{m-1}{2}$

By the paralelogram rule for sine $\sin^2 x- \sin^2 y=\sin(x-y)\sin(x+y)$ so we get that the RHS product

$P=\sin x \prod_{1 \leq j \leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi j }{m } \right) \right)=(-1)^{\frac{m-1}{2}}\prod_{0 \leq j \leq m-1}\sin (x+\frac{j\pi}{m})=$

$=(-1)^{\frac{m-1}{2}}2^{-(m-1)}\sin mx$ by the classic product formula, so we are done!

(the product formula is obtained by taking the imaginary part of both sides in $e^{2imx}-1=\Pi_{k=0,..m-1} {(e^{2ix}-e^{-\frac{2\pi ik}{m}})}$)

Answer 2

Too long for a comment: Look up Chebyshev polynomial of second kind. They are literally what you are dealing with: $$U_n(\cos x) = \dfrac{\sin((n+1)x)}{\sin x}.$$

Your attempt at induction basically reduces it to an equivalent problem that uses the first kind of Chebyshev polynomials, so I would not be fixated on inductive approach.

Answer 3

This answer can be seen as supplement of @Conrads answer providing some more details.

We start with the right-hand side of OPs identity. Letting $m=2k+1$ we obtain: \begin{align*} \color{blue}{(-4)^k}&\color{blue}{\prod_{j=1}^k\left(\sin^2(x)-\sin^2\left(\frac{2j\pi}{2k+1}\right)\right)}\\ &=(-4)^k\prod_{j=1}^k\left[\sin\left(x+\frac{2j\pi}{2k+1}\right)\sin\left(x-\frac{2j\pi}{2k+1}\right)\right]\tag{1}\\ &=4^k\prod_{j=1}^k\left[\sin\left(x+\frac{2j\pi}{2k+1}\right)\sin\left(\left(x-\frac{2j\pi}{2k+1}\right)-\pi\right)\right]\tag{2}\\ &=4^k\left(\prod_{j=1}^k\sin\left(x+\frac{2j\pi}{2k+1}\right)\right)\left(\prod_{j=1}^k\sin\left(x+\frac{(2k+1-2j)\pi}{2k+1}\right)\right)\tag{3}\\ &=4^k\left(\prod_{j=1}^k\sin\left(x+\frac{2j\pi}{2k+1}\right)\right)\left(\prod_{j=1}^k\sin\left(x+\frac{(2j-1)\pi}{2k+1}\right)\right)\tag{4}\\ &=4^k\left(\prod_{{j=1}\atop{j\ even}}^{2k}\sin\left(x+\frac{j\pi}{2k+1}\right)\right)\left(\prod_{{j=1}\atop{j\ odd}}^{2k}\sin\left(x+\frac{j\pi}{2k+1}\right)\right)\tag{5}\\ &\,\,\color{blue}{=4^k\prod_{j=1}^{2k}\sin\left(x+\frac{j\pi }{2k+1}\right)}\tag{6} \end{align*}

Comment:

• In (1) we recall the trigonometric addition formulas \begin{align*} \sin(x+y)&=\sin(x)\cos(y)+\cos(x)\sin(y)\\ \sin(x-y)&=\sin(x)\cos(y)-\cos(x)\sin(y) \end{align*} and get \begin{align*} \sin&(x+y)\sin(x-y)\\ &=\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\left(\sin(x)\cos(y)-\cos(x)\sin(y)\right)\\ &=\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)\\ &=\sin^2(x)\left(1-\sin^2(y)\right)-\left(1-\sin^2(x)\right)\sin^2(y)\\ &=\sin^2(x)-\sin^2(y) \end{align*}

• In (2) we use the identity $\sin(x)=\sin(\pi -x)$ and factor out $(-1)^k$ by using $\sin(x)=-\sin(-x)$.

• In (3) we use $\sin(x)=\sin(x+2\pi)$ and we split the product as preparation for the next steps.

• In (4) we change the order of the multiplication in the right-hand product $j\to k-j+1$.

• In (5) we do not change anything. We just write the index region somewhat more conveniently to better see the next step, where the products can be merged.

In order to simplify (6) we recall Euler's formula $e^{ix}=\cos(x)+i\sin(x)$. We obtain \begin{align*} \color{blue}{4^k}&\color{blue}{\prod_{j=0}^{2k}\sin\left(x+\frac{j\pi }{2k+1}\right)}\\ &=4^k\prod_{j=0}^{2k}\left[\frac{1}{2i}\left(e^{i\left(x+\frac{j\pi }{2k+1}\right)}-e^{-i\left(x+\frac{j\pi }{2k+1}\right)}\right)\right]\tag{7}\\ &=\frac{(-1)^{k+1}}{2i}\prod_{j=0}^{2k}\left[e^{-i\left(x+\frac{j\pi}{2k+1}\right)}\left(1-e^{2i\left(x+\frac{j\pi}{2k+1}\right)}\right)\right]\tag{8}\\ &=\frac{(-1)^{k+1}}{2i}e^{-i(2k+1)x}e^{-\frac{i\pi}{2k+1}\sum_{j=0}^{2k}j} \prod_{j=0}^{2k}\left(1-e^{2i\left(x+\frac{j\pi}{2k+1}\right)}\right)\tag{9}\\ &=\frac{(-1)^{k+1}}{2i}e^{-i(2k+1)x}e^{-ik\pi} \prod_{j=0}^{2k}\left(1-\left(e^{\frac{2\pi i}{2k+1}}\right)^j e^{2ix}\right)\tag{10}\\ &=\frac{(-1)}{2i}e^{-i(2k+1)x} \left(1-\left(e^{2ix}\right)^{2k+1} \right)\tag{11}\\ &=\frac{1}{2i}\left(e^{(2k+1)ix}-e^{-(2k+1)ix}\right)\\ &\,\,\color{blue}{=\sin((2k+1)x)} \end{align*} and the claim follows.

Comment:

• In (7) we use the identity $\sin(x)=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)$.

• In (8) we factor out $\left(\frac{1}{2i}\right)^{2k+1}$ from the product and within the product $e^{-i\left(x+\frac{j\pi}{2k+1}\right)}$.

• In (9) we factor out some more terms which do not depend on the index $j$.

• In (10) we use the summation formula $\sum_{j=1}^{2k}j = \frac{1}{2}(2k)(2k+1)$ , the identity $e^{ik\pi}=(-1)^k$ and we write the factor in the product in the form \begin{align*} 1-\omega ^j z \end{align*} with $\omega=e^{\frac{2\pi i}{2k+1}}$ the $(2k+1)$-st root of unity.

• In (11) we use the representation with $\omega$ the root of unity and $z=e^{2ix}$. \begin{align*} \prod_{j=0}^{2k}\left(1-z\omega^j\right)=\left(1+z+\cdots+z^{2k}\right)(1-z)=1-z^{2k+1} \end{align*}