Question Number 188376 by Rupesh123 last updated on 28/Feb/23
Answered by Frix last updated on 28/Feb/23
$$\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right) \\ $$$$\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{7}}{\mathrm{256}}\left(\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$
Commented by Rupesh123 last updated on 28/Feb/23
Correct, sir!
Commented by Rupesh123 last updated on 01/Mar/23
Please explain the trig identities
Commented by BaliramKumar last updated on 01/Mar/23
$$\mathrm{2}{sinAsinB}\:{formula} \\ $$