Question Number 32968 by diyatrivedi last updated on 08/Apr/18 Answered by MJS last updated on 08/Apr/18 $$\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta}=\mathrm{cos}\:\alpha×\mathrm{cos}\:\theta+\mathrm{sin}\:\alpha×\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\alpha\:=\mathrm{cos}\:\alpha×\mathrm{cos}\:\theta×\mathrm{sin}\:\theta+\mathrm{sin}\alpha×\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\mathrm{sin}\:\alpha=\mathrm{cos}\:\alpha×\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}+\mathrm{sin}\:\alpha×\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{2}}=\mathrm{cos}\:\alpha×\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}−\mathrm{sin}\:\alpha×\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}} \\…
Question Number 163936 by cortano1 last updated on 12/Jan/22 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ Answered by mahdipoor last updated on 12/Jan/22 $${get}\:{cos}\beta=\frac{{x}^{\mathrm{2}}…
Question Number 163926 by cortano1 last updated on 12/Jan/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\left.\begin{matrix}{\mathrm{sin}\:\theta=\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\\{\mathrm{csc}\:^{\mathrm{2}} \theta−\mathrm{tan}\:^{\mathrm{2}} \theta\:=?\:}\end{matrix}\right\} \\ $$ Answered by MJS_new last updated on 12/Jan/22 $${s}^{\mathrm{2}} =\mathrm{1}−{s} \\…
Question Number 32841 by alimudinade@gmail.com last updated on 03/Apr/18 $${A}+{B}+{C}=\mathrm{180}^{\mathrm{0}} \\ $$$$\mathrm{3}{sin}\:{A}\:+\mathrm{4}{cosB}\:=\mathrm{6} \\ $$$$\mathrm{3}{cosA}\:+\:\mathrm{4}{sin}\:{B}\:=\sqrt{\mathrm{13}} \\ $$$${sin}\:{C}=…. \\ $$ Answered by hknkrc46 last updated on 04/Apr/18…
Question Number 163798 by kdaramaths last updated on 10/Jan/22 $$ \\ $$$$\:\:\:\:{by}\:{a}\:{simple}\:{method}\:{calculate}\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}} {COS}\left({kx}\right) \\ $$$$ \\ $$$$ \\ $$ Answered…
Question Number 98191 by behi83417@gmail.com last updated on 12/Jun/20 $$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\::\left\{\boldsymbol{\mathrm{x}}\mid\:\mathrm{0}<\boldsymbol{\mathrm{x}}<\mathrm{2}\boldsymbol{\pi}\right\}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{cosec}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}−\mathrm{9}=\boldsymbol{\mathrm{cotx}} \\ $$ Answered by MJS last updated on 12/Jun/20 $$\mathrm{4cosec}^{\mathrm{2}} \:{x}\:−\mathrm{9}=\mathrm{cot}\:{x} \\…
Question Number 98149 by me2love2math last updated on 11/Jun/20 Commented by me2love2math last updated on 11/Jun/20 $${Q}\mathrm{1}\:{and}\:\mathrm{2}\:{pls} \\ $$ Answered by mr W last updated…
Question Number 32538 by GAUTHAM last updated on 27/Mar/18 Commented by prof Abdo imad last updated on 28/Mar/18 $${we}\:{have}\:{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)=\:\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow\:{sin}\left(\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:\:\Rightarrow \\ $$$${arcsin}\left(\:\:\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{12}}\:. \\…
Question Number 97994 by hardylanes last updated on 10/Jun/20 $${find}\:{all}\:{the}\:{values}\:{of}\:\theta,\:{in}\:{the}\:{interval}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$${for}\:{which}\:\mathrm{sin}\:\mathrm{3}\theta−\mathrm{sin}\:\theta=\sqrt{\mathrm{3cos}\:\mathrm{2}\theta} \\ $$ Answered by Rio Michael last updated on 10/Jun/20 $$\mathrm{i}\:\mathrm{pressume}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\theta\:−\:\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{for}\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\mathrm{2}\pi…
Question Number 32163 by jarjum last updated on 20/Mar/18 $${If}\:{the}\:{corrdinater}\:{of}\:{the}\:{verticle}\:{of}\:{an} \\ $$$${eqvilateral}\:{triangle}\:{with}\:{length}\:{x}\:{are} \\ $$$$\left({x}_{\mathrm{1}+} {y}_{\mathrm{1}} \right),\left({y}_{\mathrm{1}} +{y}_{\mathrm{2}} \right)\:{and}\:\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right)\:{then} \\ $$$$\left(\begin{vmatrix}{{x}_{\mathrm{1}} \:\:\:{y}_{\mathrm{1}} \:\:\:\mathrm{2}}\\{{x}_{\mathrm{2}} \:\:\:{y}_{\mathrm{2}}…