Chemistry, asked by TheMahakals, 10 months ago

Prove the following using complex numbers.

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

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Answered by BRAINLYADDICTOR

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

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Prove the following using complex numbers. ​

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Prove the following using complex numbers.

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