Category: Trigonometry

Question Number 20886 by tammi last updated on 06/Sep/17 $$if\sin x=m\sin y$$$$soproofthat$$$$\tan \frac{1}{2}(x-y)=\frac{m-1}{m+1}\tan \frac{1}{2}(x+y)$$ Answered by ajfour last updated on 06/Sep/17 $${m}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:\:\:\:\Rightarrow\:\:\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}=\frac{\mathrm{sin}\:{x}−\mathrm{sin}\:{y}}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}} \…

Question Number 20884 by tammi last updated on 06/Sep/17 $$2\cos \frac{\pi }{3}\cos \frac{9\pi }{13}+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13}=0$$ Answered by ajfour last updated on 06/Sep/17 $$muststartwith2\cos \frac{\pi }{13}\dots ,Then$$$$L.H.S.=\cos \left(\frac{10\pi }{13}\right)+\cos \left(\frac{8\pi }{13}\right)+$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \…

Question Number 151922 by john_santu last updated on 24/Aug/21 $$\sqrt[3]{\cos \frac{2\pi }{9}}+\sqrt[3]{\cos \frac{4\pi }{9}}-\sqrt[3]{\cos \frac{\pi }{9}}=?$$$$$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 20785 by Tinkutara last updated on 02/Sep/17 $$\mathrm{If}\mathrm{in}\mathrm{a}\mathrm{\Delta }ABC,\tan \frac{A}{2},\tan \frac{B}{2}\mathrm{and}\tan \frac{C}{2}$$$$\mathrm{are}\mathrm{in}\mathrm{HP},\mathrm{then}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}$$$$\cot \frac{B}{2}\mathrm{is}\mathrm{greater}\mathrm{than}$$$$\left(1\right)2\sqrt{3}$$$$\left(2\right)\frac{\sqrt{3}}{2}$$$$\left(3\right)\sqrt{3}$$$$\left(4\right)\frac{1}{\sqrt{3}}$$ Answered…

Question Number 20703 by tammi last updated on 01/Sep/17 $$\frac{\sin \alpha +\sin 3\alpha +\sin 5\alpha }{\cos \alpha +\cos 3\alpha +\cos 5\alpha }=\tan 3\alpha$$ Commented by myintkhaing last updated on 01/Sep/17 $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{2\sin 3\alpha \cos 2\alpha +\sin 3\alpha }{2\cos 3\alpha \cos 2\alpha +\cos 3\alpha }=\frac{\sin 3\alpha }{\cos 3\alpha }=\tan 3\alpha$$$$\mathrm{proved}$$ Answered…