Question Number 170902 by mathlove last updated on 03/Jun/22 $${prove}\:{that} \\ $$$${sec}\frac{\mathrm{2}\pi}{\mathrm{7}}+{sec}\frac{\mathrm{4}\pi}{\mathrm{7}}+{sec}\frac{\mathrm{8}\pi}{\mathrm{7}}=−\mathrm{4} \\ $$$$ \\ $$ Commented by som(math1967) last updated on 03/Jun/22 $${Q}\:{no}\:\mathrm{165558} \\…
Question Number 170872 by mathlove last updated on 02/Jun/22 Answered by aleks041103 last updated on 04/Jun/22 $$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}{sin}\left(\frac{{k}\pi}{{n}}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{e}^{\frac{{k}\pi{i}}{{n}}} −{e}^{−\frac{{k}\pi{i}}{{n}}} }{\mathrm{2}{i}}= \\ $$$$=\frac{\left(\underset{{k}=\mathrm{1}}…
Question Number 105290 by bemath last updated on 27/Jul/20 $$\mathcal{G}{iven}\:\frac{\mathrm{sin}\:\mathrm{2}{a}−\mathrm{sin}\:\mathrm{2}{b}}{\mathrm{cos}\:\mathrm{2}{a}+\mathrm{cos}\:\mathrm{2}{b}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\left({a}−{b}\right)\: \\ $$ Answered by bobhans last updated on 27/Jul/20 $$\frac{\mathrm{2cos}\:\left(\frac{\mathrm{2}{a}+\mathrm{2}{b}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}{a}−\mathrm{2}{b}}{\mathrm{2}}\right)}{\mathrm{2cos}\:\left(\frac{\mathrm{2}{a}+\mathrm{2}{b}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}{a}−\mathrm{2}{b}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\left({a}−{b}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\rightarrow\begin{cases}{\mathrm{sin}\:\left({a}−{b}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}}\\{\mathrm{cos}\:\left({a}−{b}\right)\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}}\end{cases} \\…
Question Number 105285 by bemath last updated on 27/Jul/20 $${simplify}\:\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{cot}\:\theta}\:+\:\frac{\mathrm{cos}\:\theta}{\mathrm{1}+\:\mathrm{tan}\:\theta}\:? \\ $$ Commented by bemath last updated on 27/Jul/20 $${thank}\:{you} \\ $$ Commented by som(math1967)…
Question Number 105243 by bemath last updated on 27/Jul/20 $$\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}\:+\:\mathrm{sin}\:\mathrm{2}{y}\:=\:\frac{\mathrm{4}}{\mathrm{9}}}\\{\mathrm{cos}\:\left({x}−{y}\right)\:=\:\mathrm{1}−\mathrm{sin}\:\left({x}+{y}\right)}\end{cases} \\ $$$$\mathrm{0}\:<\:{x}\:<\:\frac{\pi}{\mathrm{2}}\:;\:\mathrm{0}\:<\:{y}\:<\:\frac{\pi}{\mathrm{2}} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left({x}+{y}\right) \\ $$$$\left({a}\right)\:−\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\left({b}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\left({c}\right)\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left({d}\right)\:\frac{\mathrm{2}}{\mathrm{9}}\:\:\:\:\left({e}\right)\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered by bramlex last updated…
Question Number 170752 by balirampatel last updated on 30/May/22 $${Solve}:−\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 30/May/22 $$\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$$$\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}}\\\hline\end{array} \\ $$$$\:\:\frac{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)+\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)}{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)−\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)} \\…
Question Number 170662 by MikeH last updated on 28/May/22 $$\mathrm{Given}\:\mathrm{that}\: \\ $$$${I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\:{I}_{{n}+\mathrm{2}} \:=\:\left(\frac{{I}_{{n}+\mathrm{1}} }{{I}_{{n}+\mathrm{4}} }\right){I}_{{n}} \\…
Question Number 105112 by bemath last updated on 26/Jul/20 $$\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{0}\: \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\: \\ $$ Answered by john santu last updated on 28/Jul/20 $$\Leftrightarrow\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)+\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{0}…
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Question Number 39486 by ajfour last updated on 06/Jul/18 $$\mathrm{sin}\:\theta=\mathrm{sin}\:\alpha\mathrm{sin}\:\left(\frac{\theta+\alpha}{\mathrm{2}}\right) \\ $$$${Express}\:\theta\:{explicitly}\:{in}\:{terms}\:{of}\:\alpha. \\ $$ Commented by math khazana by abdo last updated on 07/Jul/18 $$\Rightarrow\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)={sin}\alpha\left({sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)+\right.…