Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$${lets}\:{take}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$$${sin}\mathrm{4}\theta={sin}\frac{\mathrm{4}\pi}{\mathrm{3}}={sin}\left(\pi+\frac{\pi}{\mathrm{3}}\right)=−{sin}\frac{\pi}{\mathrm{3}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}} \\ $$$$=−{sin}\frac{\pi}{\mathrm{3}}+\mathrm{4}{sin}\frac{\mathrm{2}\pi}{\mathrm{3}}{cos}\frac{\pi}{\mathrm{6}} \\ $$$$=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}{sin}\left(\pi−\frac{\pi}{\mathrm{3}}\right)×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}{sin}\frac{\pi}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{4}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{3}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\neq−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\therefore−{sin}\theta+\mathrm{4}{sin}\mathrm{2}\theta{cos}\frac{\theta}{\mathrm{2}}\neq{sin}\mathrm{4}\theta\:{for}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$$${so}\:{we}\:{can}\:{not}\:{prove}\:{the}\:{given}\:{question}. \\ $$$${please}\:{Recheck}\:{the}\:{question} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com