Question Number 206362 by MATHEMATICSAM last updated on 12/Apr/24
$$\mathrm{sin}\frac{\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\:? \\ $$
Answered by Skabetix last updated on 12/Apr/24
$$=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$
Answered by Berbere last updated on 13/Apr/24
$${sin}\left(\frac{\pi}{\mathrm{7}}\right)={sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right);{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)={sin}\left(\frac{\mathrm{5}\pi}{\mathrm{7}}\right);{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)={sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right) \\ $$$${Z}^{\mathrm{7}} −\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\prod}}\left({Z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{7}}} \right) \\ $$$$\frac{{Z}^{\mathrm{7}} −\mathrm{1}}{{Z}−\mathrm{1}}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\left({Z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{7}}} \right)=\Sigma{z}^{{k}} \\ $$$${z}=\mathrm{1},{z}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{7}}} =−{e}^{\frac{{ik}\pi}{\mathrm{7}}} .\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{7}}\right) \\ $$$$\Rightarrow\left(−\mathrm{2}{i}\right)^{\mathrm{6}} .{e}^{\mathrm{3}{i}\pi} \Pi{sin}\left(\frac{{k}\pi}{\mathrm{7}}\right)=\mathrm{6}\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{7}}\right)=\frac{\mathrm{7}}{\mathrm{64}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{7}}\right){sin}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{7}}\right)=\sqrt{\frac{\mathrm{7}}{\mathrm{64}}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$$$ \\ $$