Question

If $k=\sin \frac{\mathrm{\pi }}{18}\sin \frac{5\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18},$ then the numerical value of k is __________.

Answer 1

$$\begin{gathered} \mathrm{Given},k=\sin \frac{\mathrm{\pi }}{18}\sin \frac{5\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{2}\left[2\sin \frac{\mathrm{\pi }}{18}\sin \frac{5\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18}\right] \\ \left[\mathrm{using}\mathrm{identity}:2\sin A\sin B=\cos (A-B)-\cos (A+B)\right] \\ =\frac{1}{2}\left\{\left[\cos \left(\frac{5\mathrm{\pi }}{18}-\frac{\mathrm{\pi }}{18}\right)-\cos \left(\frac{5\mathrm{\pi }}{18}+\frac{\mathrm{\pi }}{18}\right)\right]\sin \frac{7\mathrm{\pi }}{18}\right\} \\ =\frac{1}{2}\left\{\left[\cos \frac{4\mathrm{\pi }}{18}-\cos \frac{6\mathrm{\pi }}{18}\right]\sin \frac{7\mathrm{\pi }}{18}\right\} \\ =\frac{1}{2}\left[\cos \frac{4\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18}-\cos \frac{6\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18}\right] \\ =\frac{1}{2}\cos \frac{4\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18}-\frac{1}{2}\cos \frac{\mathrm{\pi }}{3}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{2}\left[\frac{1}{2}\left(2\cos \frac{4\mathrm{\pi }}{18}\sin \frac{7\mathrm{\pi }}{18}\right)\right]-\frac{1}{2}\times \frac{1}{2}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{2}\times \frac{1}{2}\left[\sin \left(\frac{4\mathrm{\pi }}{18}+\frac{7\mathrm{\pi }}{18}\right)+\sin \left(\frac{7\mathrm{\pi }}{18}-\frac{4\mathrm{\pi }}{18}\right)\right]-\frac{1}{4}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{4}\left[\sin \frac{11\mathrm{\pi }}{18}+\sin \frac{3\mathrm{\pi }}{18}\right]-\frac{1}{4}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{4}\left[\sin \left(\mathrm{\pi }-\frac{7\mathrm{\pi }}{18}\right)+\sin \frac{\mathrm{\pi }}{6}\right]-\frac{1}{4}\sin \frac{7\mathrm{\pi }}{18} \\ =\frac{1}{4}\sin \frac{7\mathrm{\pi }}{18}+\frac{1}{4}\sin \frac{\mathrm{\pi }}{6}-\frac{1}{4}\sin \frac{7\mathrm{\pi }}{18}\left(\mathrm{since}\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\sin \theta \right) \\ =\frac{1}{4}\times \sin \frac{\mathrm{\pi }}{6} \\ =\frac{1}{4}\times \frac{1}{2} \\ =\frac{1}{8} \\ \mathrm{i}.\mathrm{e}.k=\frac{1}{8} \end{gathered}$$