Question Number 105692 by bramlex last updated on 31/Jul/20 $$\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{7}{x}\right)=\mathrm{1} \\ $$ Answered by john santu last updated on 31/Jul/20 $$\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5}{x}\right)+\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{7}{x}\right)=\mathrm{1}…
Question Number 105650 by bemath last updated on 30/Jul/20 $$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)+\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$${find}\:{x} \\ $$ Answered by bramlex last updated on 30/Jul/20 $$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)=\frac{\pi}{\mathrm{6}}−\mathrm{cos}^{−\mathrm{1}}…
Question Number 105619 by bobhans last updated on 30/Jul/20 $${find}\:\mathcal{G}{eneral}\:{solution}\:\mathrm{cot}\:{x}+\mathrm{cot}\:\mathrm{2}{x}+\mathrm{cot3}{x}=\:\mathrm{0} \\ $$$$ \\ $$ Answered by bemath last updated on 30/Jul/20 $$\Leftrightarrow\mathrm{cot}\:\left({x}\right)+\frac{\mathrm{cot}\:^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{\mathrm{2cot}\:\left({x}\right)}+\frac{\mathrm{3cot}\:\left({x}\right)−\mathrm{cot}\:^{\mathrm{3}} \left({x}\right)}{\mathrm{1}−\mathrm{3cot}\:^{\mathrm{2}} \left({x}\right)}=\:\mathrm{0}…
Question Number 105575 by Algoritm last updated on 30/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170968 by 281981 last updated on 05/Jun/22 Commented by mr W last updated on 05/Jun/22 $$\mathrm{tan}^{\mathrm{2}} \:\left({x}+{y}\right)+\mathrm{cos}^{\mathrm{2}} \:\left({x}+{y}\right)−\mathrm{1}+\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\left({x}+{y}\right)−\mathrm{sin}^{\mathrm{2}} \:\left({x}+{y}\right)+\left({y}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 105366 by bemath last updated on 28/Jul/20 Answered by $@y@m last updated on 28/Jul/20 $$\mathrm{15}.\:\frac{\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}+\mathrm{1}\right)+\mathrm{tan}\:{A}\left(\mathrm{sec}\:{A}−\mathrm{1}\right)}{\mathrm{sec}^{\mathrm{2}} \:{A}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2tan}\:{A}\mathrm{sec}\:{A}}{\mathrm{tan}\:^{\mathrm{2}} {A}} \\ $$$$=\mathrm{2cosec}\:{A} \\ $$…
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)…