Category: Trigonometry

Question Number 24948 by adityapratap2585@gmail.com last updated on 29/Nov/17 $$\mathrm{Assuming}\mathrm{that}\mathrm{the}{\mathrm{moon}}^{\prime }\mathrm{s}\mathrm{diameter}$$$$\mathrm{subtends}\mathrm{and}\mathrm{angle}\left(1/2\right)°\mathrm{at}\mathrm{the}\mathrm{eye}$$$$\mathrm{of}\mathrm{an}\mathrm{observer},\mathrm{find}\mathrm{how}\mathrm{far}\mathrm{from}\mathrm{the}$$$$\mathrm{eye}\mathrm{of}\mathrm{a}\mathrm{coin}\mathrm{of}10\mathrm{cm}\mathrm{diameter}\mathrm{must}$$$$\mathrm{be}\mathrm{held}\mathrm{so}\mathrm{as}\mathrm{just}\mathrm{to}\mathrm{hide}\mathrm{moon}?$$ Commented by adityapratap2585@gmail.com last updated…

Question Number 155989 by cortano last updated on 07/Oct/21 $$5\sin {}^{2}2\mathrm{x}+8\cos {}^3\mathrm{x}=8\cos \mathrm{x}$$$$\frac{3\pi }{2}⩽\mathrm{x}⩽2\pi$$ Commented by john_santu last updated on 07/Oct/21 $$\:{x}=\frac{\mathrm{3}\pi}{\mathrm{2}},\:\mathrm{2}\pi−\mathrm{arccos}\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{2}\pi \…

Question Number 155898 by Tawa11 last updated on 05/Oct/21 $$\mathrm{Show}\mathrm{that}{\tan }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{4}$$ Answered by immortel last updated on 05/Oct/21 $${L}={tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)=\frac{{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)+{tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right){tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)} \…

Question Number 24554 by tawa tawa last updated on 20/Nov/17 $$\mathrm{Show}\mathrm{that}:{\tan }^{-1}\left(\frac{\mathrm{p}}{\mathrm{p}+2\mathrm{q}}\right)+{\tan }^{-1}\left(\frac{\mathrm{p}}{\mathrm{p}+\mathrm{q}}\right)=\frac{\pi }{2}$$ Commented by mrW1 last updated on 21/Nov/17 $${that}^{\prime }snottrue!$$$${let}'{s}\:{say}\:{p}={q}=\mathrm{1}…

Question Number 89951 by john santu last updated on 20/Apr/20 $$\log {}_2\left(\sin (x+\frac{5\pi }{12})\right)+\log {}_2\left(\sin (x+\frac{\pi }{12})\right)=-1$$ Commented by jagoll last updated on 20/Apr/20 $$\Rightarrow \sin (\mathrm{x}+\frac{5\pi }{12})\sin (\mathrm{x}+\frac{\pi }{12})=\frac{1}{2}$$$$\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{12}}\right)−\mathrm{cos}\:\left(\mathrm{2x}+\frac{\mathrm{6}\pi}{\mathrm{12}}\right)\:=\:\mathrm{1}…