Trigonometry Questions

Question Number 144169 by gsk2684 last updated on 22/Jun/21

$$\mathrm{please}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\$$$$\mathrm{sin}\:\frac{\mathrm{2}\Pi}{\mathrm{7}}+\mathrm{sin}\:\frac{\mathrm{4}\Pi}{\mathrm{7}}+\mathrm{sin}\:\frac{\mathrm{8}\Pi}{\mathrm{7}} \\$$

Commented by justtry last updated on 23/Jun/21

Answered by mindispower last updated on 23/Jun/21

$${sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right){sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right) \\$$$$={sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right) \\$$$$=\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{7}}\right)\left({cos}\left(\frac{\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right) \\$$$$\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{7}}\right){cos}\left(\frac{\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right) \\$$$$=\mathrm{4}{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{14}}\right){cos}\left(\frac{\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right)={S} \\$$$${S}^{\mathrm{2}} =−\mathrm{16}{cos}\left(\frac{\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{13}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{11}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{5}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{9}\pi}{\mathrm{14}}\right) \\$$$${just}\:{using}\:{cos}\left(\pi−{x}\right)=−{cos}\left({x}\right) \\$$$$\\$$$$\\$$$${z}^{\mathrm{7}} +\mathrm{1}=\mathrm{0}\Rightarrow{z}={e}^{{i}\left(\frac{\pi+\mathrm{2}{k}\pi}{\mathrm{7}}\right)} ,{k}\in\left\{\mathrm{0},\mathrm{6}\right\} \\$$$${z}^{\mathrm{7}} +\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\prod}}\left({z}−{e}^{\frac{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \right) \\$$$$\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{6}} −{z}^{\mathrm{5}} +{z}^{\mathrm{4}} −{z}^{\mathrm{3}} +{z}^{\mathrm{2}} −{z}+\mathrm{1}\right)=\left({z}+\mathrm{1}\right)\underset{{k}=\mathrm{0},{k}\neq\mathrm{3}} {\overset{\mathrm{6}} {\prod}}\left({z}−{e}^{{i}\frac{\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \right) \\$$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\left(−{z}\right)^{{k}} =\underset{{k}=\mathrm{0},{k}\neq\mathrm{3}} {\overset{\mathrm{6}} {\prod}}\left({z}−{e}^{\frac{\left(\mathrm{1}+\mathrm{2}{k}\right){i}\pi}{\mathrm{7}}} \right) \\$$$${z}=−\mathrm{1} \\$$$$\Rightarrow\mathrm{7}=\Pi\left(\mathrm{1}+{e}^{\frac{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \right)=\underset{{k}=\mathrm{0},{k}\neq\mathrm{3}} {\overset{\mathrm{6}} {\prod}}.{e}^{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\frac{\pi}{\mathrm{14}}} \left({e}^{\frac{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{14}}} +{e}^{\frac{−{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi}{\mathrm{14}}} \right) \\$$$$\Rightarrow\mathrm{7}=\mathrm{2}^{\mathrm{6}} .{e}^{\frac{{i}\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{9}+\mathrm{11}+\mathrm{13}\right)\pi}{\mathrm{14}}} .\underset{{k}=,{k}\neq\mathrm{3}} {\overset{\mathrm{6}} {\prod}}{cos}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{14}}\pi\underset{={S}'} {\right)} \\$$$$\mathrm{7}=\mathrm{2}^{\mathrm{6}} .{e}^{{i}\left(\mathrm{3}\pi\right)} .{S} \\$$$${S}'=−\frac{\mathrm{7}}{\mathrm{64}} \\$$$${S}'=−\frac{{S}^{\mathrm{2}} }{\mathrm{16}} \\$$$$\Rightarrow−\frac{{S}^{\mathrm{2}} }{\mathrm{16}}=−\frac{\mathrm{7}}{\mathrm{64}}\Rightarrow{S}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{4}},{S}\in\left\{\frac{\sqrt{\mathrm{7}}}{\mathrm{2}},−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right. \\$$$$\frac{\pi}{\mathrm{14}},\frac{\mathrm{3}\pi}{\mathrm{14}},\frac{\mathrm{5}\pi}{\mathrm{14}}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\Rightarrow{S}>\mathrm{0}\right.\right. \\$$$${S}={sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\$$$$\\$$