Chemistry, asked by TheMahakals, 8 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)}

Answers

Answered by BRAINLYADDICTOR
87

\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 Theopekaaleader
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the correct answer I will be

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