(b) We have $k = \sin \frac{\pi }{{18}}\sin \frac{{5\pi }}{{18}}\sin \frac{{7\pi }}{{18}}$
$ = \cos \left( {\frac{\pi }{2} – \frac{\pi }{{18}}} \right)\cos \left( {\frac{\pi }{2} – \frac{{5\pi }}{{18}}} \right)\cos \left( {\frac{\pi }{2} – \frac{{7\pi }}{{18}}} \right)$
$ = \cos \frac{\pi }{9}\cos \frac{{2\pi }}{9}\cos \frac{{4\pi }}{9} = \frac{{\sin {2^3}\frac{\pi }{9}}}{{{2^3}\sin \frac{\pi }{9}}} = \frac{{\sin \frac{{8\pi }}{9}}}{{8\sin \frac{\pi }{9}}}$
$ = \frac{{\sin \left( {\pi – \frac{\pi }{9}} \right)}}{{8\sin \frac{\pi }{9}}} = \frac{1}{8}$.