Category: Trigonometry

Question Number 48711 by somil last updated on 27/Nov/18 Commented by mr W last updated on 27/Nov/18 $$A+B+C=\pi$$$$A+B=\pi -C$$$$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$$$$\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \…

Question Number 114108 by bemath last updated on 17/Sep/20 $$provethat2{\tan }^{-1}\left(\frac{2}{3}\right)={\sin }^{-1}\left(\frac{12}{13}\right)$$ Answered by bobhans last updated on 17/Sep/20 $$let{\tan }^{-1}\left(\frac{2}{3}\right)=x\to \tan x=\frac{2}{3}$$$${and}\:$$\{\begin{array}{l}\sin x=\frac{2}{\sqrt{13}}\\ \cos x=\frac{3}{\sqrt{13}}\end{array}$$…

Question Number 179550 by cortano1 last updated on 30/Oct/22 $$16\cos {}^5\theta -\cos 5\theta =\frac{5}{2}$$$$0<\theta <2\pi$$$$\theta =?$$ Answered by greougoury555 last updated on 30/Oct/22 $$\:\:$$\{\begin{array}{l}\cos \theta =c\\ \sin \theta =s\end{array}$$\Rightarrow\mathrm{cos}\:\mathrm{5}\theta=\left(\mathrm{4}{c}^{\mathrm{3}}…

Question Number 48360 by maxmathsup by imad last updated on 22/Nov/18 $$calculate{cos}^4\left(\frac{\pi }{8}\right)+{cos}^4\left(\frac{3\pi }{8}\right)+{cos}^4\left(\frac{5\pi }{8}\right)+{cos}^4\left(\frac{7\pi }{8}\right)$$ Commented by Abdo msup. last updated on…

Question Number 113769 by faysal last updated on 15/Sep/20 $$provethat,\tan \left(7\frac{1}{2}\right)°=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$$ Answered by Dwaipayan Shikari last updated on 15/Sep/20 $$tan\left(\frac{\pi }{24}\right)=\frac{sin\frac{\pi }{24}}{cos\frac{\pi }{24}}=\frac{2{sin}^2\frac{\pi }{24}}{2sin\frac{\pi }{24}cos\frac{\pi }{24}}=\frac{1-cos\frac{\pi }{12}}{sin\frac{\pi }{12}}$$$${cos}\frac{\pi}{\mathrm{12}}={cos}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \…