Question Number 208252 by efronzo1 last updated on 09/Jun/24
$$\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{22}}\right)\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{20}\pi}{\mathrm{21}}\right)=? \\ $$
Answered by som(math1967) last updated on 09/Jun/24
$$\:{let}\:\frac{\pi}{\mathrm{21}}={x} \\ $$$${cos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{16}{xcos}\mathrm{20}{x} \\ $$$${cosxcos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{5}{x} \\ $$$$\left[\mathrm{21}{x}=\pi\Rightarrow{cos}\mathrm{20}{x}=−{cosx},{cos}\mathrm{16}{x}=−{cos}\mathrm{5}{x}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{sinxsin}\mathrm{5}{x}}×{sin}\mathrm{2}{xcos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xsin}\mathrm{10}{xcos}\mathrm{10}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}{sinxsin}\mathrm{5}{x}}×\mathrm{2}{sin}\mathrm{8}{xcos}\mathrm{8}{x}×\mathrm{2}{sin}\mathrm{10}{xcoss}\mathrm{10}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}{sinxsin}\mathrm{5}{x}}×{sin}\mathrm{16}{xsin}\mathrm{20}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}{sinxsin}\mathrm{5}{x}}×{sin}\mathrm{5}{x}×{sinx} \\ $$$$\left[\mathrm{21}{x}=\pi\:\Rightarrow{sin}\mathrm{16}{x}={sin}\mathrm{5}{x},{sin}\mathrm{20}{x}={sinx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}} \\ $$
Commented by efronzo1 last updated on 09/Jun/24
Commented by som(math1967) last updated on 09/Jun/24
$${yes} \\ $$
Commented by Ghisom last updated on 09/Jun/24
$$\mathrm{I}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result} \\ $$