Question

Solve: $\sin \frac{\pi }{18}\sin \frac{3\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}=\frac{1}{16}$

Answer 1

Converting the angles in radians to degrees, we get,

$=\sin \left({10}^{\circ }\right)\sin \left({30}^{\circ }\right)\sin \left({50}^{\circ }\right)\sin \left({70}^{\circ }\right)$

$=\frac{\sin \left({10}^{\circ }\right)\sin \left({50}^{\circ }\right)\sin \left({70}^{\circ }\right)}{2}$ ------ $\sin \left({30}^{\circ }\right)=\frac{1}{2}$

We know that, $\cos ({90}^{\circ }-x)=\sin \left(x\right)$

$=\frac{\cos \left({80}^{\circ }\right)\cos \left({40}^{\circ }\right)\cos \left({20}^{\circ }\right)}{2}$

$=\frac{\left(\cos \right({80}^{\circ }+{40}^{\circ })+\cos ({80}^{\circ }-{40}^{\circ }\left)\right)\cos \left({20}^{\circ }\right)}{4}$

$=\frac{(\frac{-1}{2}+\cos ({40}^{\circ }\left)\right)\cos \left({20}^{\circ }\right)}{4}$

$=\frac{\left(\cos \right({40}^{\circ }\left)\cos \right({20}^{\circ })-\frac{\cos \left({20}^{\circ }\right)}{2}}{4}$

$=\frac{2\cos \left({40}^{\circ }\right)\cos \left({20}^{\circ }\right)-\cos \left({20}^{\circ }\right)}{8}$

$=\frac{\cos ({40}^{\circ }+{20}^{\circ })+\cos ({40}^{\circ }-{20}^{\circ })-\cos \left({20}^{\circ }\right)}{8}$

$=\frac{\frac{1}{2}+\cos \left({20}^{\circ }\right)-\cos \left({20}^{\circ }\right)}{8}$

$=\frac{\frac{1}{2}}{8}$

$=\frac{1}{16}$