Question Number 10788 by okhema last updated on 25/Feb/17 $${factorise}\:{the}\:{expression}\:{sin}\mathrm{4}{x}−{sinx} \\ $$ Answered by argemiroQR last updated on 26/Feb/17 $${sin}\left(\mathrm{3}{x}+{x}\right)−{sinx}= \\ $$$${sin}\mathrm{3}{xcosx}+{cos}\mathrm{3}{xsinx}−{sinx}= \\ $$$${cosx}\left(\mathrm{3}{sinx}−\mathrm{4}{sin}^{\mathrm{3}} {x}\right)+\left({cos}^{\mathrm{3}}…
Question Number 10750 by Joel576 last updated on 24/Feb/17 $$\mathrm{Exact}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{sin}\:\mathrm{9}°\:=\:… \\ $$ Answered by ridwan balatif last updated on 24/Feb/17 $$\mathrm{first}\:\mathrm{we}\:\mathrm{must}\:\mathrm{know}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{18}^{\mathrm{o}} \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{18}^{\mathrm{o}}…
Question Number 10746 by okhema last updated on 24/Feb/17 $$\left.{i}\right){express}\:{the}\:{function}\:{f}\left(\theta\right)={sin}\theta\:+\:{cos}\theta\:{in}\:{the}\:{form}\:{rsin}\left(\theta+\alpha\right),\:{r}>\mathrm{0}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\left.{ii}\right){hence}\:{find}\:{the}\:{maximum}\:{value}\:{of}\:{f}\:{and} \\ $$$${the}\:{smallest}\:{non}−{negative}\:{value}\:{of}\:\theta\:{at}\:{which}\:{it}\:{occurs}. \\ $$ Answered by mrW1 last updated on 24/Feb/17 $$\left.{i}\right) \\…
Question Number 10743 by okhema last updated on 24/Feb/17 $${show}\:{that}\:\mathrm{sec}\:^{\mathrm{2}} \theta=\frac{\mathrm{cosec}\:\theta}{\mathrm{cosec}\:\theta−\mathrm{sin}\:} \\ $$ Commented by ridwan balatif last updated on 24/Feb/17 $$\frac{\mathrm{cosec}\theta}{\mathrm{cosec}\theta−\mathrm{sin}\theta}=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}}{\mathrm{sin}\theta}−\mathrm{sin}\theta}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{sin}\theta}}{\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}}…
Question Number 10744 by okhema last updated on 24/Feb/17 $${hence}\:{or}\:{otherwise},{solve}\:{the}\:{equation}\:\frac{\mathrm{cosec}\:\theta}{\mathrm{cosec}\:\theta−\mathrm{sin}\:\theta}=\frac{\mathrm{4}}{\mathrm{3}}\:{for}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\Pi \\ $$ Answered by malwaan last updated on 24/Feb/17 $${cox}\:\theta=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{30}°\:\:\boldsymbol{{or}}\:\boldsymbol{\theta}=\mathrm{180}−\mathrm{30}=\mathrm{150}° \\ $$$${or}\:\theta=\mathrm{180}+\mathrm{30}=\mathrm{210}°\:{or}\:\theta=−\mathrm{30}° \\…
Question Number 76270 by Master last updated on 25/Dec/19 Commented by Master last updated on 25/Dec/19 $$\left.\mathrm{A}\left.\right)\:\mathrm{sin6}°+\mathrm{cos6}°\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}\right)\mathrm{sin12}°+\mathrm{cos12}° \\ $$$$\left.\mathrm{C}\left.\right)\mathrm{sin18}°+\mathrm{cos18}°\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{sin24}°+\mathrm{cos24}° \\ $$ Commented by MJS last…
Question Number 76265 by A8;15: last updated on 25/Dec/19 Answered by Tanmay chaudhury last updated on 26/Dec/19 $${S}={cos}\alpha+{cos}\mathrm{2}\alpha+{cos}\mathrm{3}\alpha+…+{cosn}\alpha \\ $$$${multiply}\:{each}\:{term}\:{by}\:\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:{and}\:{apply}\:{trigo} \\ $$$${formula}\:{and}\:{adding}\:{them} \\ $$$$\mathrm{2}{cos}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)={sin}\left(\frac{\mathrm{3}\alpha}{\mathrm{2}}\right)−{sin}\left(\frac{\alpha}{\mathrm{2}}\right) \\…
Question Number 10695 by okhema last updated on 23/Feb/17 $${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}} \\…