# Simple Complex numbers

$\large \dfrac2{1 - e^{i \pi /5}} = \dfrac{e^{i \cdot 2\pi /5}}{\sin \frac\pi{10}}$

Prove the equation above.

Note by TheKiz Zer
2 years, 4 months ago

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$e^{\frac{i\pi}{5}}=\cos \frac{\pi}{5}+i\sin \frac{\pi}{5}\\ 1-\cos \frac{\pi}{5}=2\sin^2\frac{\pi}{10}\\ \sin \frac{\pi}{5}=2\sin \frac{\pi}{10}\cos \frac{\pi}{10}$ such that given expression simplifies to:- $\dfrac{1}{\sin \frac{\pi}{10}(\cos (\frac{-2\pi}{5})+i\sin (\frac{-2\pi}{5}))}$ $\large =\dfrac{e^{i \cdot 2\pi /5}}{\sin \frac\pi{10}}$

- 2 years, 4 months ago