Previous in Relation and Functions Next in Relation and Functions
Question Number 42508 by maxmathsup by imad last updated on 26/Aug/18
$${find}\:{the}\:{value}\:{of}\: \\ $$$${A}\:={cos}\left(\frac{\pi}{\mathrm{5}}\right).{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:{and}\:{B}\:={sin}\left(\frac{\pi}{\mathrm{5}}\right){sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right). \\ $$
Answered by math1967 last updated on 27/Aug/18
$${let}\frac{\pi}{\mathrm{5}}=\theta\:\:\therefore\mathrm{4}\theta=\pi−\theta \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} {sin}\theta}×{sin}\mathrm{4}\theta×{cos}\mathrm{4}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{sin}\theta}×{sin}\left(\pi−\theta\right)×{cos}\mathrm{4}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×{cos}\left(\pi−\theta\right)=−\frac{\mathrm{1}}{\mathrm{4}}{cos}\frac{\pi}{\mathrm{5}}=−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{16}} \\ $$
Commented by rahul 19 last updated on 27/Aug/18
Nice!
Answered by math1967 last updated on 27/Aug/18
![B=sinθsin2θsin4θ [let θ=(π/5)] B=sinθsin2θsin(π−θ) =sin^2 θsin2θ=(1/(16))(10−2(√5) )×(1/4)(√(10+2(√5)))](Q42524.png)
$${B}={sin}\theta{sin}\mathrm{2}\theta{sin}\mathrm{4}\theta\:\:\left[{let}\:\theta=\frac{\pi}{\mathrm{5}}\right] \\ $$$${B}={sin}\theta{sin}\mathrm{2}\theta{sin}\left(\pi−\theta\right) \\ $$$$={sin}^{\mathrm{2}} \theta{sin}\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}\:\right)×\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$