Category: Trigonometry

Question Number 15641 by tawa tawa last updated on 12/Jun/17 $$\mathrm{Show}\mathrm{that}.$$$$\sqrt{3}\left(\mathrm{cosec}20\right)-\sec 20=4$$ Answered by mrW1 last updated on 12/Jun/17 $$\frac{\sqrt{3}}{\sin 20}-\frac{1}{\cos 20}=\frac{\sqrt{3}\cos 20-1\sin 20}{\sin 20\cos 20}$$$$=\frac{\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}\right)}{\mathrm{sin}\:\mathrm{40}}…

Question Number 81149 by TawaTawa last updated on 09/Feb/20 Answered by mind is power last updated on 09/Feb/20 $$tan\left(B\right)=\frac{AC}{AB}$$$$tg(B-\theta )=\frac{AD}{AB}\Rightarrow \frac{tg\left(B\right)}{tg(B-\theta )}=\frac{AC}{AD}=4$$$$\Rightarrow tg\left(B\right)=4tg(B-\theta )\Rightarrow tg\left(B\right)=4\frac{tg\left(B\right)-tg\left(\theta \right)}{1+tg\left(B\right)tg\left(\theta \right)}$$$$\Rightarrow\mathrm{3}{tg}\left({B}\right)={tg}\left(\theta\right)\left({tg}^{\mathrm{2}}…

Question Number 15555 by Tinkutara last updated on 11/Jun/17 $$\mathrm{Prove}\mathrm{that}$$$$\frac{\cos 8x-\cos 7x}{1+2\cos 5x}=\cos 3x-\cos 2x$$ Answered by ajfour last updated on 11/Jun/17 $$considering$$$${f}\left({x}\right)=\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{5}{x}\right)\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}\right) \…

Question Number 81057 by mathocean1 last updated on 09/Feb/20 $$solve\mathrm{in}[-\pi ;\pi ]$$$$\left(E\right):sin3x=-sin2x$$ Commented by jagoll last updated on 09/Feb/20 $$\sin 3x+\sin 2x=0$$$$\mathrm{2sin}\:\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0} \…

Question Number 80998 by Power last updated on 08/Feb/20 Commented by mr W last updated on 08/Feb/20 $$justasforalotofyourotherquestions,$$$$youalsoneedacalculatorforthisone.$$ Answered by jagoll…