Write the value of sin
Question:

Write the value of $\sin \frac{\pi}{15} \sin \frac{4 \pi}{15} \sin \frac{3 \pi}{10}$

Solution:

$\frac{\pi}{15}=12^{\circ}, \frac{4 \pi}{15}=48^{\circ}, \frac{3 \pi}{10}=54^{\circ}$

$\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}$

$=\frac{1}{2}\left[2 \sin 12^{\circ} \sin 48^{\circ}\right] \sin 54^{\circ}$

$=\frac{1}{2}\left[\cos \left(12^{\circ}-48^{\circ}\right)-\cos \left(12^{\circ}+48^{\circ}\right)\right] \sin 54^{\circ}$

$=\frac{1}{2}\left[\cos \left(-36^{\circ}\right)-\cos 60^{\circ}\right] \sin 54^{\circ}$

$=\frac{1}{2} \sin 54^{\circ}\left[\cos 36^{\circ}-\frac{1}{2}\right]$

$=\frac{1}{2}\left[\sin \left(90^{\circ}-36^{\circ}\right) \cos 36^{\circ}\right]-\frac{1}{4} \sin \left(90^{\circ}-36^{\circ}\right)$

$=\frac{1}{2} \cos ^{2} 36^{\circ}-\frac{1}{4} \cos 36^{\circ}$

$=\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)^{2}-\left(\frac{\sqrt{5}+1}{16}\right) \quad\left[\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right]$

$=\frac{1}{2}\left(\frac{5+1+2 \sqrt{5}}{16}\right)-\left(\frac{\sqrt{5}+1}{16}\right)$

$=\frac{6+2 \sqrt{5}}{32}-\frac{\sqrt{5}+1}{16}$

$=\frac{6+2 \sqrt{5}-2 \sqrt{5}-2}{32}$

$=\frac{4}{32}$

$=\frac{1}{8}$