Category: Trigonometry

Question Number 65345 by KanhAshish last updated on 28/Jul/19 $$\mathrm{find}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}{\sin }^{2018}\mathrm{x}+{\cos }^{2019}\mathrm{x}?$$ Commented by mr W last updated on 28/Jul/19 $${maximum}\:{of}\:\mathrm{sin}^{{m}} \:{x}+\mathrm{cos}^{{n}} \:{x}\:{is}\:\mathrm{1}.…

Question Number 130632 by bramlexs22 last updated on 27/Jan/21 Answered by EDWIN88 last updated on 27/Jan/21 $$\cos 4095°=\cos (360°\times 11+135°)$$$$=-\frac{\sqrt{2}}{2}$$$$then{\sin }^{-1}\left(\cos 4095°\right)=\sin (-\frac{\sqrt{2}}{2})$$$$\:=−\frac{\pi}{\mathrm{4}} \…

Question Number 130581 by bramlexs22 last updated on 27/Jan/21 $$\mathrm{If}{\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3\mathrm{are}\mathrm{in}\mathrm{AP}\mathrm{and}\mathrm{d}\mathrm{is}\mathrm{the}\mathrm{common}$$$$\mathrm{difference}\mathrm{then}{\tan }^{-1}\left(\frac{\mathrm{d}}{1+{\mathrm{a}}_1{\mathrm{a}}_2}\right)+{\tan }^{-1}\left(\frac{\mathrm{d}}{1+{\mathrm{a}}_2{\mathrm{a}}_3}\right)=?$$ Answered by…

Question Number 130469 by liberty last updated on 26/Jan/21 $$P=\cos \frac{\pi }{15}.\cos \frac{2\pi }{15}.\cos \frac{3\pi }{15}.\cos \frac{4\pi }{15}.\cos \frac{5\pi }{15}.\cos \frac{6\pi }{15}.\cos \frac{7\pi }{15}$$ Answered by EDWIN88 last updated on 26/Jan/21 $$2\mathrm{psin}\frac{\pi }{15}=\sin \frac{2\pi }{15}.\cos \frac{2\pi }{15}.\cos \frac{3\pi }{15}.\cos \frac{4\pi }{15}.\cos \frac{\pi }{3}.\cos \frac{6\pi }{15}.\cos \frac{7\pi }{15}$$$$4\mathrm{psin}\frac{\pi }{15}=\sin \frac{4\pi }{15}.\cos \frac{3\pi }{15}.\cos \frac{4\pi }{15}.\frac{1}{2}.\cos \frac{6\pi }{15}.\cos \frac{7\pi }{15}$$$$\mathrm{8psin}\:\frac{\pi}{\mathrm{15}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \…

Question Number 64773 by ankan0 last updated on 21/Jul/19 $${\sin }^{-1}(1/\sqrt{5)}+{\cot }^{-1}3$$ Answered by Kunal12588 last updated on 21/Jul/19 $$\mathrm{cot}^{−\mathrm{1}} \left({a}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right)…