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

$$\mathrm{Prove}\mathrm{that}:$$$$2\sin \frac{1}{2}\theta \cos \frac{3}{2}\theta +2\sin \frac{5}{2}\theta$$$$+2\sin \frac{3}{2}\theta +2\sin \frac{3}{2}\theta \cos \frac{7}{2}\theta$$$$=\sin 4\theta +\sin 5\theta$$

Answered by Kunal12588 last updated on 03/Mar/19

$$2sin\frac{1}{2}\theta cos\frac{3}{2}\theta +2sin\frac{5}{2}\theta +2sin\frac{3}{2}\theta +2sin\frac{3}{2}\theta cos\frac{7}{2}\theta$$$$=sin\frac{3\theta +\theta }{2}-sin\frac{3\theta -\theta }{2}+2(sin\frac{5}{2}\theta +sin\frac{3}{2}\theta )+sin\frac{3\theta +7\theta }{2}-sin\frac{7\theta -3\theta }{2}$$$$=sin2\theta -sin\theta +sin5\theta -sin2\theta +4sin\frac{(5\theta +3\theta )/2}{2}cos\frac{(5\theta -3\theta )/2}{2}$$$$=sin5\theta -sin\theta +4sin2\theta cos\frac{\theta }{2}$$$$soifwecanprove-sin\theta +4sin2\theta cos\frac{\theta }{2}=sin4\theta$$$$wecanprovethegivenquestion$$

Commented by Kunal12588 last updated on 03/Mar/19

$$letstake\theta =\frac{\pi }{3}$$$$sin4\theta =sin\frac{4\pi }{3}=sin(\pi +\frac{\pi }{3})=-sin\frac{\pi }{3}=-\frac{\sqrt{3}}{2}$$$$-sin\theta +4sin2\theta cos\frac{\theta }{2}$$$$=-sin\frac{\pi }{3}+4sin\frac{2\pi }{3}cos\frac{\pi }{6}$$$$=-\frac{\sqrt{3}}{2}+4sin(\pi -\frac{\pi }{3})\times \frac{\sqrt{3}}{2}$$$$=\frac{-\sqrt{3}}{2}+4sin\frac{\pi }{3}\times \frac{\sqrt{3}}{2}=\frac{-\sqrt{3}}{2}+4\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$$$$=3-\frac{\sqrt{3}}{2}\ne -\frac{\sqrt{3}}{2}$$$$\therefore -sin\theta +4sin2\theta cos\frac{\theta }{2}\ne sin4\theta for\theta =\frac{\pi }{3}$$$$sowecannotprovethegivenquestion.$$$$pleaseRecheckthequestion$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com