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

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:- \[\dfrac{1}{\sin \frac{\pi}{10}(\cos (\frac{-2\pi}{5})+i\sin (\frac{-2\pi}{5}))}\] \[\large =\dfrac{e^{i \cdot 2\pi /5}}{\sin \frac\pi{10}} \]

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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