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Question Number 55704 by gunawan last updated on 03/Mar/19
Prove that: 2 sin (1/2)θcos (3/2)θ+2sin (5/2)θ +2 sin (3/2)θ+2sin (3/2)θcos (7/2)θ =sin 4θ+sin 5θ
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\theta\mathrm{cos}\:\frac{\mathrm{3}}{\mathrm{2}}\theta+\mathrm{2sin}\:\frac{\mathrm{5}}{\mathrm{2}}\theta\: \\ $$$$+\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{3}}{\mathrm{2}}\theta+\mathrm{2sin}\:\frac{\mathrm{3}}{\mathrm{2}}\theta\mathrm{cos}\:\frac{\mathrm{7}}{\mathrm{2}}\theta \\ $$$$=\mathrm{sin}\:\mathrm{4}\theta+\mathrm{sin}\:\mathrm{5}\theta \\ $$
Answered by Kunal12588 last updated on 03/Mar/19
2sin(1/2)θ cos(3/2)θ +2 sin(5/2)θ + 2sin (3/2)θ + 2sin(3/2)θ cos(7/2)θ =sin((3θ+θ)/2)−sin((3θ−θ)/2)+2(sin(5/2)θ+sin(3/2)θ)+sin((3θ+7θ)/2)−sin((7θ−3θ)/2) =sin2θ−sinθ+sin5θ−sin2θ+4sin(((5θ+3θ)/2)/2)cos(((5θ−3θ)/2)/2) =sin5θ−sinθ+4sin2θcos(θ/2) so if we can prove −sinθ+4sin2θcos(θ/2)=sin4θ we can prove the given question
$$\mathrm{2}{sin}\frac{\mathrm{1}}{\mathrm{2}}\theta\:{cos}\frac{\mathrm{3}}{\mathrm{2}}\theta\:+\mathrm{2}\:{sin}\frac{\mathrm{5}}{\mathrm{2}}\theta\:+\:\mathrm{2}{sin}\:\frac{\mathrm{3}}{\mathrm{2}}\theta\:+\:\mathrm{2}{sin}\frac{\mathrm{3}}{\mathrm{2}}\theta\:{cos}\frac{\mathrm{7}}{\mathrm{2}}\theta \\ $$$$={sin}\frac{\mathrm{3}\theta+\theta}{\mathrm{2}}−{sin}\frac{\mathrm{3}\theta−\theta}{\mathrm{2}}+\mathrm{2}\left({sin}\frac{\mathrm{5}}{\mathrm{2}}\theta+{sin}\frac{\mathrm{3}}{\mathrm{2}}\theta\right)+{sin}\frac{\mathrm{3}\theta+\mathrm{7}\theta}{\mathrm{2}}−{sin}\frac{\mathrm{7}\theta−\mathrm{3}\theta}{\mathrm{2}} \\ $$$$={sin}\mathrm{2}\theta−{sin}\theta+{sin}\mathrm{5}\theta−{sin}\mathrm{2}\theta+\mathrm{4}{sin}\frac{\left(\mathrm{5}\theta+\mathrm{3}\theta\right)/\mathrm{2}}{\mathrm{2}}{cos}\frac{\left(\mathrm{5}\theta−\mathrm{3}\theta\right)/\mathrm{2}}{\mathrm{2}} \\ $$$$={sin}\mathrm{5}\theta−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}} \\ $$$${so}\:{if}\:{we}\:{can}\:{prove}\:−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}}={sin}\mathrm{4}\theta \\ $$$${we}\:{can}\:{prove}\:{the}\:{given}\:{question} \\ $$
Commented by Kunal12588 last updated on 03/Mar/19
lets take θ=(π/3) sin4θ=sin((4π)/3)=sin(π+(π/3))=−sin(π/3)=−((√3)/2) −sinθ+4sin2θcos(θ/2) =−sin(π/3)+4sin((2π)/3)cos(π/6) =−((√3)/2)+4sin(π−(π/3))×((√3)/2) =((−(√3))/2)+4sin(π/3)×((√3)/2)=((−(√3))/2)+4×((√3)/2)×((√3)/2) =3−((√3)/2) ≠−((√3)/2) ∴−sinθ+4sin2θcos(θ/2)≠sin4θ for θ=(π/3) so we can not prove the given question. please Recheck the question
$${lets}\:{take}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$$${sin}\mathrm{4}\theta={sin}\frac{\mathrm{4}\pi}{\mathrm{3}}={sin}\left(\pi+\frac{\pi}{\mathrm{3}}\right)=−{sin}\frac{\pi}{\mathrm{3}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}} \\ $$$$=−{sin}\frac{\pi}{\mathrm{3}}+\mathrm{4}{sin}\frac{\mathrm{2}\pi}{\mathrm{3}}{cos}\frac{\pi}{\mathrm{6}} \\ $$$$=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}{sin}\left(\pi−\frac{\pi}{\mathrm{3}}\right)×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}{sin}\frac{\pi}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{3}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\neq−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\therefore−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}}\neq{sin}\mathrm{4}\theta\:{for}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$$${so}\:{we}\:{can}\:{not}\:{prove}\:{the}\:{given}\:{question}. \\ $$$${please}\:{Recheck}\:{the}\:{question} \\ $$

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