Question Number 9800 by richard last updated on 05/Jan/17 $$\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:\mathrm{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:\:? \\ $$ Commented by FilupSmith last updated on 05/Jan/17 $${tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}={tg}^{\mathrm{2}} \frac{\pi}{\mathrm{4}} \\ $$$$\:…
Question Number 140805 by jlewis last updated on 12/May/21 $${find}\:{x}\:\mathrm{2cosh}\:\mathrm{2}{x}+\mathrm{10sinh}\:\mathrm{2}{x}=\mathrm{5} \\ $$ Answered by Ar Brandon last updated on 12/May/21 $$\mathrm{2cosh2x}+\mathrm{10sinh2x}=\mathrm{5} \\ $$$$\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} +\mathrm{5}\left(\mathrm{e}^{\mathrm{2x}}…
Question Number 140803 by jlewis last updated on 12/May/21 Answered by Ar Brandon last updated on 12/May/21 $$\mathrm{tany}=\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:,\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}=\mathrm{sec}^{\mathrm{2}} \mathrm{y} \\ $$$$\Rightarrow\mathrm{cosy}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}}}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}}…
Question Number 9709 by tawakalitu last updated on 28/Dec/16 $$\mathrm{A}\:\mathrm{vertical}\:\mathrm{pole}\:\mathrm{3m}\:\mathrm{high}\:\mathrm{is}\:\mathrm{2m}\:\mathrm{south}\:\mathrm{of}\:\mathrm{a}\:\mathrm{wall} \\ $$$$\mathrm{which}\:\mathrm{runs}\:\mathrm{directly}\:\mathrm{east}\:\mathrm{and}\:\mathrm{west}.\:\mathrm{The}\:\mathrm{sun} \\ $$$$\mathrm{is}\:\mathrm{south}\:\mathrm{west}\:\mathrm{at}\:\mathrm{an}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{35}°. \\ $$$$\mathrm{Find}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{centimeter}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{shadow}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pole}\:\mathrm{on}\:\mathrm{the}\:\mathrm{wall}. \\ $$ Answered by mrW last updated…
Question Number 9699 by ridwan balatif last updated on 26/Dec/16 Commented by ridwan balatif last updated on 26/Dec/16 $$\mathrm{how}\:\mathrm{many}\:\mathrm{x}\:\mathrm{can}\:\mathrm{fill}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{cos3x}+\mathrm{tan3x}\right)\left(\mathrm{cos3x}−\mathrm{tan3x}\right)=\mathrm{1}\:\mathrm{for} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi,\:\mathrm{x}\neq\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2k}\pi}{\mathrm{3}}\:\mathrm{and}\:\mathrm{k}\:\mathrm{is}\:\left(\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…\right)\: \\ $$$$\mathrm{is}\:….?…
Question Number 9704 by tawakalitu last updated on 26/Dec/16 Answered by geovane10math last updated on 26/Dec/16 Commented by tawakalitu last updated on 28/Dec/16 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\…
Question Number 140767 by EDWIN88 last updated on 12/May/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\::\:\mathrm{2cot}\:^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{csc}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}\:=\:\mathrm{0}\: \\ $$ Answered by Ar Brandon last updated on 12/May/21 $$\mathrm{2cot}^{\mathrm{2}} \mathrm{x}+\mathrm{csc}^{\mathrm{2}} \mathrm{x}−\mathrm{2}=\mathrm{0}…
Question Number 9690 by tawakalitu last updated on 24/Dec/16 Commented by ridwan balatif last updated on 25/Dec/16 $$\mathrm{10}\:\mathrm{is}\:\mathrm{radius},\mathrm{right}? \\ $$ Commented by tawakalitu last updated…
Question Number 140746 by 676597498 last updated on 12/May/21 Answered by Ar Brandon last updated on 12/May/21 $$\left(\mathrm{i}\right)\:\left(\mathrm{1}+\mathrm{3w}\right)\left(\mathrm{1}+\mathrm{3w}^{\mathrm{2}} \right)=\mathrm{1}+\mathrm{3}\left(\mathrm{w}^{\mathrm{2}} +\mathrm{w}\right)+\mathrm{9w}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{3}\left(−\mathrm{w}^{\mathrm{3}} \right)+\mathrm{9}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{3}+\mathrm{9}=\mathrm{7} \\ $$$$…
Question Number 75208 by vishalbhardwaj last updated on 08/Dec/19 $$\:\mathrm{Prove}\:\mathrm{that}\::\:\mathrm{cos}\:\mathrm{18}^{\mathrm{0}} −\mathrm{sin}\:\mathrm{18}^{\mathrm{0}} \:=\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{2} \\ $$ Answered by Kunal12588 last updated on 08/Dec/19 $${cos}\:\mathrm{18}°−{sin}\mathrm{18}° \\ $$$$=\sqrt{\mathrm{2}}{sin}\left(\mathrm{45}°−\mathrm{18}°\right) \\…