Question

Prove the following using complex numbers.

$\displaystyle \frac{\sin((2n+1)\theta)}{(2n+1)\sin(\theta)} = \prod_{r=1}^n \left( 1-\frac{\sin^2(\theta)}{\sin^2\left(\frac{r\pi}{(2n+1)}\right)} \right )$

Answer 1

Step-by-step explanation:

left off:

sin(2kθ)2sinθ+cos((2k+1)θ)

sin⁡(2kθ)2sin⁡θ+cos⁡((2k+1)θ)

Let x=2kθx=2kθ. Then the expression is

sinx2sinθ+cos(x+θ)

sin⁡x2sin⁡θ+cos⁡(x+θ)

Angle sum identity gives

sinx2sinθ+cosxcosθ−sinxsinθ

sin⁡x2sin⁡θ+cos⁡xcos⁡θ−sin⁡xsin⁡θ

=sinx+2cosxcosθsinθ−2sinxsin2θ2sinθ.

=sin⁡x+2cos⁡xcos⁡θsin⁡θ−2sin⁡xsin2⁡θ2sin⁡θ.

Factoring, we get

=(1−2sin2θ)sinx+(2cosθsinθ)cosx2sinθ.

=(1−2sin2⁡θ)sin⁡x+(2cos⁡θsin⁡θ)cos⁡x2sin⁡θ.

We use the double angle formula and angle sum formula in reverse:

=cos2θsinx+sin2θcosx2sinθ

=cos⁡2θsin⁡x+sin⁡2θcos⁡x2sin⁡θ

=sin(x+2θ)2sinθ

=sin⁡(x+2θ)2sin⁡θ

=sin(2kθ+2θ)2sinθ

=sin⁡(2kθ+2θ)2sin⁡θ

=sin(2(k+1)θ)2sinθ.

=sin⁡(2(k+1)θ)2sin⁡θ.

This completes the inductive step.

Answer 2

Answer :

$\textsf{RHS} &\equiv \prod_{r=1}^n \left( 1-\frac{\sin^2(\theta)}{\sin^2\left(\frac{r\pi}{2n+1}\right)}\right)\\&=\frac{4^n}{m}\prod_{r=1}^n \left( \sin^2\left(r\phi\right)-\sin^2(\theta) \right) \qquad \text{Here: } \phi \equiv \frac{\pi}{2n+1} = \frac12\arg(\xi) \\ &=\frac{4^n}{m}\prod_{r=1}^n \left( \sin\left(r\phi - \theta \right) \sin\left(r\phi + \theta \right) \right) \\&=\frac{4^n}{m}\prod_{r=1}^n \left( \frac{(e^{i(r\phi - \theta)}-e^{-i(r\phi - \theta)})}{2i}\frac{(e^{i(r\phi +\theta)}-e^{-i(r\phi + \theta)})}{2i} \right) \\&=\color{orange}{\frac{4^n}{m}\frac{1}{(-4)^n}}\prod_{r=1}^n (w^r/z-z/w^r)(w^rz-w^{-r}/z) \qquad \text{Here: } w \equiv e^{i\phi}\,\,(w^2=\xi) \\&=\color{orange}{\frac{(-1)^n}{m}}\prod_{r=1}^n \left(\frac{\color{red}{-z}}{\color{red}{-z}}(w^{r}/z-z/w^r)\frac{\color{blue}{w^r}}{1}\right)\left(\frac{1}{\color{blue}{w^r}}(w^rz-w^{-r}/z)\frac{\color{green}{z}}{\color{green}{z}}\right) \\&=\frac{(-1)^n}{m}\prod_{r=1}^n \left(\frac{(-w^{2r}+z^2)}{\color{red}{-z}}\right)\left(\frac{(z^2-w^{-2r})}{\color{green}{z}}\right) \\&=\frac{(-1)^n}{m}(\color{red}{-z})^{-n}\color{green}{z}^{-n}\prod_{r=1}^n (z^2-\xi^2)\prod_{r=1}^n(z^2-\xi^{-2}) \\&=\frac{1}{\color{red}{z}^{n}\color{green}{z}^{n}m}\prod_{r=1}^n (u-\xi^r)\prod_{r=1}^n(u-\xi^{-r}) \\&=\frac{1}{u^nm}\prod_{r=1}^{m-1} (u-\xi^r)\\ &=\textsf{LHS}$