# sin-4-pi-16-sin-4-3pi-16-sin-4-5pi-16-sin-4-7pi-16-

Question Number 131492 by Ar Brandon last updated on 05/Feb/21
$$\mathrm{sin}^{\mathrm{4}} \frac{\pi}{\mathrm{16}}+\mathrm{sin}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{16}}+\mathrm{sin}^{\mathrm{4}} \frac{\mathrm{5}\pi}{\mathrm{16}}+\mathrm{sin}^{\mathrm{4}} \frac{\mathrm{7}\pi}{\mathrm{16}} \\$$
Answered by Dwaipayan Shikari last updated on 05/Feb/21
$${sin}\frac{\pi}{\mathrm{16}}={cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{16}}\right)={cos}\left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)\:\:\:\:{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)={cos}\left(\frac{\mathrm{5}\pi}{\mathrm{16}}\right) \\$$$$\Phi={cos}^{\mathrm{4}} \frac{\pi}{\mathrm{16}}+{sin}^{\mathrm{4}} \frac{\pi}{\mathrm{16}}+{sin}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{16}}+{cos}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{16}} \\$$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\$$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+{sin}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\$$
Commented by Dwaipayan Shikari last updated on 05/Feb/21
$${It}\:{will}\:{be}\:{even}\:{nicer}\:{if}\:{you}\:{take}\:{the}\:{value}\:{of}\:{sin}\left(\frac{\pi}{\mathrm{16}}\right)\:'{s}\: \\$$$${Separately}\::\left(\:\:\::\right) \\$$$${sin}\left(\frac{\pi}{\mathrm{16}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\:\:…. \\$$
Commented by Ar Brandon last updated on 05/Feb/21
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Answered by Ar Brandon last updated on 05/Feb/21