Question Number 222193 by MrGaster last updated on 20/Jun/25
$$\mathrm{Prove}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right)}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4sin}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}=\mathrm{4}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\ $$
Answered by som(math1967) last updated on 20/Jun/25
$$\:\frac{\mathrm{1}}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\frac{{sin}\frac{\pi}{\mathrm{2}}}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\frac{\mathrm{2}{sin}\frac{\pi}{\mathrm{4}}{cos}\frac{\pi}{\mathrm{4}}}{{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)} \\ $$$$=\frac{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{8}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right){cos}\left(\frac{\pi}{\mathrm{4}}\right)}{{cos}\left(\frac{\pi}{\mathrm{8}}\right)} \\ $$$$=\mathrm{2}.\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{8}}\right){cos}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{2}\left\{{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}\:−{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\mathrm{2}\left({cos}\frac{\pi}{\mathrm{8}}\:−{sin}\frac{\pi}{\mathrm{8}}\right) \\ $$