Category: Trigonometry

Question Number 197346 by Amidip last updated on 14/Sep/23 Answered by som(math1967) last updated on 14/Sep/23 $$\frac{a+b}{a-b}=\frac{tan(\theta +\varphi )}{tan(\theta -\varphi )}$$$$\frac{a+b}{a-b}=\frac{sin(\theta +\varphi )cos(\theta -\varphi )}{sin(\theta -\varphi )cos(\theta +\varphi )}$$$$\frac{2a}{2b}=\frac{sin2\theta }{sin2\varphi }\left[usingcomponendo\mathrm{\&}dividendo\right]$$$$asin2\varphi =bsin2\theta$$$${a}^{\mathrm{2}}…

Question Number 197073 by Abdullahrussell last updated on 07/Sep/23 Answered by Frix last updated on 07/Sep/23 $$\sin \frac{\pi }{7}\sin \frac{2\pi }{7}\sin \frac{3\pi }{7}=x$$$$\mathrm{Use}\mathrm{trigonometric}\mathrm{formulas}\mathrm{to}\mathrm{get}$$$$\frac{1}{4}(\cos \frac{3\pi }{14}+\cos \frac{\pi }{14}-\sin \frac{\pi }{7})=x\left(1\right)$$$$\mathrm{Now}\mathrm{comes}\mathrm{the}“{\mathrm{trick}}^{″}$$$$\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)^{\mathrm{3}}…

Question Number 196886 by Amidip last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $$tan(\alpha +\beta )=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }=\frac{p}{q-1}$$$$\Rightarrow \frac{p}{q-1}=\frac{{sin}^2(\alpha +\beta )}{sin(\alpha +\beta )cos(\alpha +\beta )}$$$$\Rightarrow psin(\alpha +\beta )cos(\alpha +\beta )=(q-1){sin}^2(\alpha +\beta ))$$$$\Rightarrow{sin}^{\mathrm{2}}…

Question Number 196883 by Amidip last updated on 02/Sep/23 Answered by som(math1967) last updated on 02/Sep/23 $${\sin }^{-1}x+{\sin }^{-1}y=\pi -{\sin }^{-1}z$$$$\Rightarrow{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\mathrm{sin}^{−\mathrm{1}}…

Question Number 196755 by Amidip last updated on 31/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com