Question Number 114272 by bemath last updated on 18/Sep/20 $${find}\:{solution}\:{set}\:{of}\:{equation}\: \\ $$$$\mathrm{cos}\:\mathrm{3}{x}\:=\:−\mathrm{cos}\:{x}\:,\:{x}\in\left(\mathrm{0},\mathrm{3}\pi\right) \\ $$ Commented by malwaan last updated on 18/Sep/20 $$\left({i}\right)\:{cos}\:\mathrm{3}{x}=\:{cos}\:\left(\pi−{x}\right) \\ $$$$\Rightarrow\mathrm{3}{x}=\:\pm\left(\pi−{x}\right)+\mathrm{2}{k}\pi \\…
Question Number 48711 by somil last updated on 27/Nov/18 Commented by mr W last updated on 27/Nov/18 $${A}+{B}+{C}=\pi \\ $$$${A}+{B}=\pi−{C} \\ $$$$\frac{{A}+{B}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\…
Question Number 114122 by bemath last updated on 17/Sep/20 $${find}\:{the}\:{value}\:\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{7}}{\mathrm{13}}\right)\right) \\ $$ Answered by mr W last updated on 17/Sep/20 $$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{13}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{7}}{\:\sqrt{\mathrm{218}}}=\mathrm{cos}^{−\mathrm{1}}…
Question Number 114108 by bemath last updated on 17/Sep/20 $${prove}\:{that}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$ Answered by bobhans last updated on 17/Sep/20 $${let}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:{x}\:\rightarrow\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${and}\:\begin{cases}{\mathrm{sin}\:{x}=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}}\\{\mathrm{cos}\:{x}\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}}\end{cases}…
Question Number 114100 by faysal last updated on 17/Sep/20 $$ \\ $$$${If}\:\mathrm{cos}\:=\frac{{a}\mathrm{cos}\:\alpha−{b}}{{a}−{b}\mathrm{cos}\:\alpha},\:{prove}\:{that}, \\ $$$$\frac{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\theta}{\:\sqrt{{a}+{b}}}=\frac{\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\alpha}{\:\sqrt{{a}+{b}}} \\ $$ Commented by Henri Boucatchou last updated on 17/Sep/20 $${cos}?=\frac{{acos}\alpha….}{}…
Question Number 179550 by cortano1 last updated on 30/Oct/22 $$\:\:\:\mathrm{16cos}\:^{\mathrm{5}} \theta−\mathrm{cos}\:\mathrm{5}\theta\:=\:\frac{\mathrm{5}}{\mathrm{2}}\: \\ $$$$\:\:\:\mathrm{0}<\theta<\mathrm{2}\pi \\ $$$$\:\:\:\theta\:=? \\ $$ Answered by greougoury555 last updated on 30/Oct/22 $$\:\:\begin{cases}{\mathrm{cos}\:\theta=\:{c}}\\{\mathrm{sin}\:\theta\:=\:{s}}\end{cases}\Rightarrow\mathrm{cos}\:\mathrm{5}\theta=\left(\mathrm{4}{c}^{\mathrm{3}}…
Question Number 179432 by cortano1 last updated on 29/Oct/22 Commented by som(math1967) last updated on 29/Oct/22 $${x}+\mathrm{72}=\mathrm{108} \\ $$$$\Rightarrow{x}=\mathrm{36} \\ $$ Commented by CElcedricjunior last…
Question Number 48360 by maxmathsup by imad last updated on 22/Nov/18 $${calculate}\:{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:+{cos}^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\:+{cos}^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$ Commented by Abdo msup. last updated on…
Question Number 113781 by Ar Brandon last updated on 15/Sep/20 Answered by Dwaipayan Shikari last updated on 15/Sep/20 $${sin}\frac{\pi}{\mathrm{12}}{sin}\frac{\mathrm{7}\pi}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\frac{\pi}{\mathrm{2}}−{cos}\frac{\mathrm{8}\pi}{\mathrm{12}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Answered by Dwaipayan Shikari…
Question Number 113769 by faysal last updated on 15/Sep/20 $${prove}\:{that},\:\mathrm{tan}\:\left(\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\right)°=\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\mathrm{2} \\ $$ Answered by Dwaipayan Shikari last updated on 15/Sep/20 $${tan}\left(\frac{\pi}{\mathrm{24}}\right)=\frac{{sin}\frac{\pi}{\mathrm{24}}}{{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{24}}}{\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}}=\frac{\mathrm{1}−{cos}\frac{\pi}{\mathrm{12}}}{{sin}\frac{\pi}{\mathrm{12}}} \\ $$$${cos}\frac{\pi}{\mathrm{12}}={cos}\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\…