Question Number 192536 by cortano12 last updated on 20/May/23 $$\:\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$ Answered by Tomal last updated on 20/May/23 $$\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$$$\left.{R}=\frac{\left(\mathrm{3}×\mathrm{0}.\mathrm{0872}\right)+\left(\mathrm{4}×\mathrm{0}.\mathrm{996}\right)−\left(\mathrm{5}×\mathrm{0}.\mathrm{53}\right)+\left(\mathrm{35}\:\underbrace{\frown}\right.}{}\mathrm{2}×\mathrm{0}.\mathrm{974}\right) \\ $$ Terms…
Question Number 192537 by cortano12 last updated on 20/May/23 $$\:\begin{cases}{\mathrm{tan}\:\left(\alpha+\mathrm{2}\beta\right)=\sqrt{\mathrm{1}+\mathrm{2k}}}\\{\mathrm{tan}\:^{\mathrm{2}} \left(\alpha+\beta\right)\left\{\mathrm{1}+\mathrm{k}\:\mathrm{tan}\:^{\mathrm{2}} \beta\right\}=\mathrm{k}+\mathrm{tan}\:^{\mathrm{2}} \beta}\end{cases} \\ $$$$\:\mathrm{Find}\:\mathrm{cot}\:\mathrm{2}\beta\:. \\ $$ Answered by a.lgnaoui last updated on 20/May/23 $$\:\:\:\:\mathrm{tan}\:\left[\left(\alpha+\beta\right)+\beta\right]=\sqrt{\mathrm{1}+\mathrm{2k}}\:\:\:\:\:\:\:\:\:\:\:…
Question Number 192497 by cortano12 last updated on 19/May/23 $$\:\:\mathrm{tan}\:\mathrm{66}°+\mathrm{tan}\:\mathrm{12}°=\sqrt{\mathrm{3}}\:+\mathrm{tan}\:\mathrm{x} \\ $$$$\:\:\mathrm{x}=? \\ $$ Answered by Tomal last updated on 19/May/23 $$\:\:\mathrm{tan}\:\mathrm{66}°+\mathrm{tan}\:\mathrm{12}°=\sqrt{\mathrm{3}}\:+\mathrm{tan}\:\mathrm{x} \\ $$$$\Rightarrow\mathrm{2}.\mathrm{246}\:+\:\mathrm{0}.\mathrm{213}=\sqrt{\mathrm{3}\:}+\mathrm{tan}\:\mathrm{x} \\…
Question Number 126954 by bramlexs22 last updated on 25/Dec/20 $$\:\:\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{17}\pi}{\mathrm{18}}\right)=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Dec/20 $$\mathrm{2}{cos}\left(\frac{\mathrm{12}\pi}{\mathrm{36}}\right){cos}\left(\frac{\pi}{\mathrm{18}}\right)+{cos}\left(\frac{\mathrm{17}\pi}{\mathrm{18}}\right) \\ $$$$={cos}\left(\frac{\pi}{\mathrm{18}}\right)+{cos}\left(\frac{\mathrm{17}\pi}{\mathrm{18}}\right)=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{8}\pi}{\mathrm{18}}\right)=\mathrm{0} \\ $$…
Question Number 61412 by Tawa1 last updated on 02/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{multinomial}\:\mathrm{coefficient}:\:\:\:\:\:\begin{pmatrix}{\:\:\:\:\:\:\:\:\:\mathrm{9}}\\{\mathrm{3},\:\:\mathrm{5},\:\:\mathrm{1},\:\:\mathrm{0}}\end{pmatrix} \\ $$ Commented by mr W last updated on 02/Jun/19 Commented by mr W last…
Question Number 192346 by Rupesh123 last updated on 15/May/23 Answered by som(math1967) last updated on 15/May/23 $$\left(\frac{\mathrm{2}{cos}\mathrm{30}{cos}\mathrm{20}+{cos}\mathrm{70}+{cos}\left(\mathrm{180}−\mathrm{70}\right)}{{cos}\mathrm{20}}\right)^{\mathrm{8}} \\ $$$$=\left(\frac{\sqrt{\mathrm{3}}{cos}\mathrm{20}+\mathrm{cos}\:\mathrm{70}−\mathrm{cos}\:\mathrm{70}}{{cos}\mathrm{20}}\right)^{\mathrm{8}} \\ $$$$=\left(\frac{\sqrt{\mathrm{3}}{cos}\mathrm{20}}{{cos}\mathrm{20}}\right)^{\mathrm{8}} \\ $$$$=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{8}} =\mathrm{81} \\…
Question Number 192282 by universe last updated on 14/May/23 $$\:\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:+\:\mathrm{4sin}\:\frac{\pi}{\mathrm{9}}\:=\:? \\ $$ Answered by som(math1967) last updated on 14/May/23 $${tan}\mathrm{20}+\mathrm{4}{sin}\mathrm{20} \\ $$$$\frac{{sin}\mathrm{20}+\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{20}}{{cos}\mathrm{20}} \\…
Question Number 192278 by manxsol last updated on 13/May/23 $${S}={arctan}\left(\frac{\mathrm{2}}{\mathrm{1}^{\mathrm{2}} }\right)+{artan}\left(\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }\right)+…….. \\ $$ Commented by Frix last updated on 14/May/23 $$\mathrm{As}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{before}\:\left(\mathrm{but}\:\mathrm{it}\:\mathrm{was}\:\mathrm{deleted}\right): \\ $$$$\mathrm{I}\:\mathrm{approximated}\:\mathrm{it}\:\mathrm{using}\:\mathrm{software}\:\mathrm{and}\:\mathrm{got} \\…
Question Number 192270 by universe last updated on 13/May/23 $$\:{prove}\:{that} \\ $$$$\:\:\mathrm{sin}\:\mathrm{50}°\:+\:\mathrm{sin}\:\mathrm{10}°\:=\:\mathrm{cos}\:\mathrm{20}° \\ $$ Answered by Frix last updated on 13/May/23 $$\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{10}°\right)\:+\mathrm{sin}\:\mathrm{10}°= \\ $$$$=\mathrm{sin}\:\mathrm{60}°\:\mathrm{cos}\:\mathrm{10}°\:−\mathrm{cos60}°\:\mathrm{sin}\:\mathrm{10}°\:+\mathrm{sin}\:\mathrm{10}°\:= \\…
Question Number 61157 by Askash last updated on 29/May/19 $${if} \\ $$$${tan}\:\mathrm{5}\theta\:+\:{tan}\:\mathrm{4}\theta\:=\mathrm{1} \\ $$$${find}\:\mathrm{3}\theta \\ $$$$ \\ $$$$ \\ $$ Commented by mr W last…