Question Number 201033 by Mingma last updated on 28/Nov/23
Answered by Frix last updated on 28/Nov/23
$$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}}\:={x} \\ $$$$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{Using}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)={x}\:\bigstar \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:=\mathrm{1}−\mathrm{4}{x} \\ $$$$\left(\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)^{\mathrm{3}} =\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{Expanding}\:\&\:\mathrm{using}\:\mathrm{trig}.\:\mathrm{formulas}\:\Rightarrow \\ $$$$−\frac{\mathrm{15}}{\mathrm{4}}+\frac{\mathrm{31}}{\mathrm{4}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{Using}\:\bigstar\:\Rightarrow \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{19}{x}}{\mathrm{64}}+\frac{\mathrm{3}}{\mathrm{64}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left({x}=−\frac{\mathrm{3}}{\mathrm{8}}\wedge{x}=\mathrm{1}\:\mathrm{not}\:\mathrm{valid}\right) \\ $$
Commented by Mingma last updated on 28/Nov/23
Nice solution!
Answered by som(math1967) last updated on 28/Nov/23
$${let}\:\frac{\pi}{\mathrm{14}}=\theta\Rightarrow\mathrm{14}\theta=\pi \\ $$$$\:{sin}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\theta}×\mathrm{2}{sin}\theta{cos}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\theta}×{sin}\mathrm{2}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{cos}\theta}×\mathrm{2}{sin}\mathrm{2}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{cos}\theta}×\left({cos}\theta−{cos}\mathrm{5}\theta\right){sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×\left[\mathrm{2}{sin}\mathrm{9}\theta{cos}\theta−\mathrm{2}{sin}\mathrm{9}\theta{cos}\mathrm{5}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×\left({sin}\mathrm{10}\theta+{sin}\mathrm{8}\theta−{sin}\mathrm{14}\theta−{sin}\mathrm{4}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}\left[{sin}\left(\pi−\mathrm{4}\theta\right)+{sin}\left(\pi−\mathrm{6}\theta\right)−\mathrm{0}−{sin}\mathrm{4}\theta\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×{sin}\mathrm{6}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×{cos}\theta=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left[\:\pi=\mathrm{14}\theta\Rightarrow\frac{\pi}{\mathrm{2}}=\mathrm{7}\theta\Rightarrow\frac{\pi}{\mathrm{2}}−\theta=\mathrm{6}\theta\right. \\ $$$$\left.\therefore\:{sin}\left(\frac{\pi}{\mathrm{2}}−\theta\right)={sin}\mathrm{6}\theta\Rightarrow{cos}\theta={sin}\mathrm{6}\theta\right] \\ $$
Commented by Mingma last updated on 28/Nov/23
Nice solution!
Answered by Sutrisno last updated on 29/Nov/23
$$=\frac{\mathrm{2}.{cos}\left(\frac{\pi}{\mathrm{14}}\right){sin}\left(\frac{\pi}{\mathrm{14}}\right){sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}−\frac{\mathrm{4}\pi}{\mathrm{14}}\right){sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}+\frac{\mathrm{2}\pi}{\mathrm{14}}\right)}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right)}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{2}.{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right)}{\mathrm{4}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{2}{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right)}{\mathrm{8}{sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}+\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{14}}\right)}{\mathrm{8}{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by MathematicalUser2357 last updated on 20/Jan/24
$$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}} \\ $$$$=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\centerdot\left[\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}+\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}+\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)+\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]+\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)+\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]+\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)−\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{1}\right\}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{1}\right\} \\ $$