Category: Trigonometry

Question Number 45639 by peter frank last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $$1)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}}…

Question Number 176668 by mnjuly1970 last updated on 24/Sep/22 Answered by a.lgnaoui last updated on 25/Sep/22 $$\frac{2\pi }{9}=(\frac{3\pi }{9}-\frac{\pi }{9});\frac{4\pi }{9}=(\frac{3\pi }{9}+\frac{\pi }{9});\frac{8\pi }{9}=(\pi -\frac{\pi }{9})$$$$\cos \frac{3\pi }{9}=\cos \frac{\pi }{3}=\frac{1}{2}=\cos \left(\frac{2\pi }{9}\right)\cos \left(\frac{\pi }{9}\right)-\sin \left(\frac{2\pi }{9}\right)\cos \left(\frac{\pi }{9}\right)=(2\cos {}^2\left(\frac{\pi }{9}\right)-1)\cos \frac{\pi }{9}-2\sin {}^2\left(\frac{\pi }{9}\right)\cos \left(\frac{\pi }{9}\right)$$$$=\mathrm{2cos}\:^{\mathrm{3}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)\right)\mathrm{cos}\:\frac{\pi}{\mathrm{9}}…

Question Number 111114 by bobhans last updated on 02/Sep/20 $$4\sin 36°\cos 72°\sin 108°?$$ Answered by bemath last updated on 02/Sep/20 $$\sqrt{\mathrm{bemath}}$$$$\mathrm{we}\mathrm{want}\mathrm{to}\mathrm{compute}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\mathrm{4}\:\mathrm{sin}\:\mathrm{36}°\:\mathrm{cos}\:\mathrm{72}°\:\mathrm{sin}\:\mathrm{108}°\:. \…

Question Number 176603 by a.lgnaoui last updated on 23/Sep/22 $$looktheanser$$ Commented by a.lgnaoui last updated on 23/Sep/22 Commented by mr W last updated…

Question Number 45511 by Tawa1 last updated on 13/Oct/18 $$\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{ABC},\mathrm{\angle }\mathrm{ABC}={30}^0,\mathrm{and}\mathrm{AC}=10.\mathrm{A}\mathrm{circle}\mathrm{is}\mathrm{drawn}\mathrm{to}$$$$\mathrm{circumscribe}\mathrm{the}\mathrm{triangle}.\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}.$$ Terms of Service Privacy Policy Contact: info@tinkutara.com