Prove-that-sin-3pi-5-sin-4pi-5-1-5-2-

Question Number 168277 by Eulerian last updated on 07/Apr/22

Prove that \frac{\sin (\frac{3 π}{5})}{\sin (\frac{4 π}{5})} = \frac{1 + \sqrt{5}}{2}

Commented by cortano1 last updated on 07/Apr/22

\frac{\sin 108 °}{\sin 144 °} = \frac{\sin (90 ° + 18 °)}{\sin (180 ° - 36 °)}

= \frac{\cos 18 °}{\sin 36 °} = \frac{\cos 18 °}{2 \sin 18 ° \cos 18 °}

= \frac{1}{2 \sin 18 °} = \frac{1}{2 (\frac{\sqrt{5} - 1}{4})}

= \frac{2}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{\sqrt{5} + 1}{2}