Question Number 10695 by okhema last updated on 23/Feb/17 $${prove}\:{that}\:{tanx}−{cotx}=−\mathrm{2}{cot}\mathrm{2}{x} \\ $$ Answered by nume1114 last updated on 23/Feb/17 $$\:\:\:\:\mathrm{tan}\:{x}−\mathrm{cot}\:{x} \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\frac{\mathrm{sin}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{2}}…
Question Number 10696 by okhema last updated on 23/Feb/17 $${given}\:{that}\:{tan}\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{4}}{and}\:{that}\:{angle}\:{x}\:{is}\:{acute},\:{calculate},{without}\:{using}\:{a}\:{calculator}\:{the}\:{value}\:{of}\:{these}. \\ $$$$\left({a}\right)\:{cos}\mathrm{2}{x} \\ $$$$\left({b}\right)\:{sinx} \\ $$$$ \\ $$ Answered by mrW1 last updated on 23/Feb/17…
Question Number 10690 by Saham last updated on 22/Feb/17 Answered by sandy_suhendra last updated on 23/Feb/17 Commented by sandy_suhendra last updated on 23/Feb/17 $$\Delta\mathrm{TCB}\Rightarrow\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{x}}{\mathrm{TC}} \\…
Question Number 10653 by Saham last updated on 21/Feb/17 $$\mathrm{A}\:\mathrm{ship}\:\mathrm{leaves}\:\mathrm{a}\:\mathrm{port}\:\mathrm{P}\:\mathrm{which}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{latitude}\:\mathrm{20}°\mathrm{N}.\:\mathrm{It}\:\mathrm{sails}\:\mathrm{due}\:\mathrm{east} \\ $$$$\mathrm{through}\:\mathrm{30}°\:\mathrm{of}\:\:\mathrm{longitude}\:\mathrm{and}\:\mathrm{then}\:\mathrm{through}\:\mathrm{south}\:\mathrm{to}\:\mathrm{Q}\:\mathrm{which}\:\mathrm{lies} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{equator}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{it}\:\mathrm{has}\:\mathrm{travelled}, \\ $$$$\left(\mathrm{Take}\:\mathrm{the}\:\mathrm{cicumference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{to}\:\mathrm{be}\:\mathrm{40},\mathrm{000}\:\mathrm{km}\right). \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{return}\:\mathrm{jouney}\:\mathrm{it}\:\mathrm{sails}\:\mathrm{due}\:\mathrm{west}\:\mathrm{through}\:\mathrm{30}°\:\mathrm{of}\:\mathrm{longitude}\:\mathrm{and} \\ $$$$\mathrm{then}\:\mathrm{due}\:\mathrm{north}\:\mathrm{back}\:\mathrm{to}\:\mathrm{P}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{in}\:\mathrm{length}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{outward}\:\mathrm{and}\:\mathrm{return}\:\mathrm{jouney}\:\mathrm{is}\:\mathrm{approximately}\:\mathrm{201}\:\mathrm{kilometers}. \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\mathrm{201}\:\mathrm{km}\:\mathrm{and}\:\mathrm{taking}\:\mathrm{1}\:\mathrm{knot}\:\mathrm{to}\:\mathrm{be}\:\mathrm{1}.\mathrm{852}\:\mathrm{km}/\mathrm{hr}\:. \\…
Question Number 76086 by hejdj last updated on 23/Dec/19 $${what}\:{is}\:{minimal}\:{expression}\:{for}\:\mathrm{sin}\:\frac{\pi}{{k}}\: \\ $$$$\mathrm{cos}\:\frac{\pi}{{k}},\:\mathrm{tan}\:\frac{\pi}{{k}},\:\mathrm{cosec}\:\frac{\pi}{{k}},\mathrm{sec}\:\frac{\pi}{{k}}{and}\:\mathrm{cot}\:\frac{\pi}{{k}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10540 by FilupS last updated on 17/Feb/17 Commented by FilupS last updated on 17/Feb/17 $$\mathrm{All}\:\mathrm{side}\:\mathrm{lenghts}\:=\:{n} \\ $$$$\angle{CAE}=\theta \\ $$$$\mathrm{0}\leqslant\theta<\frac{\pi}{\mathrm{3}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{sections}…
Question Number 76075 by benjo last updated on 23/Dec/19 $$\mathrm{how}\:\mathrm{i}\:\mathrm{evaluate}\:\mathrm{sin}\left(\pi/\mathrm{7}\right)×\mathrm{sin}\left(\mathrm{2}\pi/\mathrm{7}\right)×\mathrm{sin}\left(\mathrm{3}\pi/\mathrm{7}\right)\: \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by john santuy last updated on 23/Dec/19 Terms of Service…
Question Number 10458 by paonky last updated on 10/Feb/17 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{sin}{x}\:+\:\mathrm{cos}{x}\:+\:\mathrm{sin}{x}\mathrm{cos}{x} \\ $$ Answered by mrW1 last updated on 10/Feb/17 $$\mathrm{sin}{x}\:+\mathrm{cos}\:{x} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{sin}\:{x}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{cos}\:{x}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\…
Question Number 75983 by mathocean1 last updated on 21/Dec/19 $$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{5}+\mathrm{3cos4x}\right) \\ $$ Commented by MJS last updated on 21/Dec/19 $$\mathrm{answer}\:\mathrm{is}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other} \\…
Question Number 75976 by mathocean1 last updated on 21/Dec/19 $$\left.\mathrm{h}\left.\mathrm{ello}\:\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\right]−\pi;\pi\right]\:\mathrm{and}\:\mathrm{place}\:\mathrm{solutions} \\ $$$$\mathrm{in}\:\mathrm{trigonometric}\:\mathrm{circle}. \\ $$$$\mathrm{cos}^{\mathrm{6}} \mathrm{x}+\mathrm{sin}^{\mathrm{6}} \mathrm{x}=\frac{\mathrm{3}}{\mathrm{8}}\left(\sqrt{\mathrm{3}}\mathrm{sin4x}+\frac{\mathrm{8}}{\mathrm{3}}\right) \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}… \\ $$ Commented by MJS last updated…