Category: Trigonometry

Question Number 148674 by bemath last updated on 30/Jul/21 $$\mathrm{Find}\max \mathrm{\&}\min \mathrm{value}\mathrm{of}$$$$\left(1\right)\mathrm{f}\left(\mathrm{x}\right)=\sin {}^3\mathrm{x}\sqrt{1+\cos {}^2\mathrm{x}}+\cos {}^3\mathrm{x}\sqrt{1+\sin {}^2\mathrm{x}}.$$$$\left(2\right)\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}\sqrt{1+\cos {}^2\mathrm{x}}+\cos \mathrm{x}\sqrt{1+\sin {}^2\mathrm{x}}.$$$$\mathrm{x}\in \mathbb{R}$$ Answered…

Question Number 17497 by tawa tawa last updated on 06/Jul/17 $$\mathrm{Evaluate}:\sin \left(9\right)°$$ Answered by mrW1 last updated on 06/Jul/17 $$18°+2\times 18°=90°-2\times 18°$$$$\cos (18°+2\times 18°)=\cos (90°-2\times 18°)=\sin \left(2\times 18°\right)$$$$\mathrm{cos}\:\mathrm{18}\:\mathrm{cos}\:\left(\mathrm{2}×\mathrm{18}\right)−\mathrm{sin}\:\mathrm{18}\:\mathrm{sin}\:\left(\mathrm{2}×\mathrm{18}\right)=\mathrm{sin}\:\left(\mathrm{2}×\mathrm{18}\right)…

Question Number 17435 by 786786AM last updated on 06/Jul/17 $$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}4\sin \frac{\pi }{24}\cos \frac{\pi }{12}\cos \frac{\pi }{6}.$$ Answered by alex041103 last updated on 07/Jul/17 $$\mathrm{Let}A=4sin\frac{\pi }{24}cos\frac{\pi }{12}cos\frac{\pi }{6}.$$$$\mathrm{Then}\mathrm{we}\mathrm{use}sin2\theta =2sin\theta cos\theta :$$$${A}=\mathrm{2}\left(\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}\right){cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}\:\frac{\mathrm{1}}{{cos}\frac{\pi}{\mathrm{24}}} \…

Question Number 17420 by sushmitak last updated on 05/Jul/17 $${\tan }^6\frac{\pi }{9}-{33\tan }^4\frac{\pi }{9}+{27\tan }^2\frac{\pi }{9}=?$$ Answered by Tinkutara last updated on 05/Jul/17 $$\tan 3\left(\frac{\pi }{9}\right)=\sqrt{3}$$$$\sqrt{\mathrm{3}}\:=\:\frac{\mathrm{3}\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:−\:\mathrm{tan}^{\mathrm{3}}…

Question Number 17348 by Tinkutara last updated on 04/Jul/17 $$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{points}\mathrm{in}(-\mathrm{\infty },\mathrm{\infty })\mathrm{for}$$$$\mathrm{which}{x}^2-x\sin x-\cos x=0\mathrm{is}$$$$\left(1\right)6$$$$\left(2\right)4$$$$\left(3\right)2$$$$\left(4\right)0$$ Answered by…