Question Number 82768 by Lontum Hans last updated on 24/Feb/20 Commented by mr W last updated on 24/Feb/20 $${if}\:{there}\:{is}\:{no}\:{gravity}\:{force},\:{the}\:{pendulum} \\ $$$${will}\:{not}\:{move}\:{at}\:{all},\:{i}.{e}.\:{its}\:{period}\:{is} \\ $$$$\infty.\:{so}\:{we}\:{know}:\:{the}\:{weaker}\:{the}\:{gravity}, \\ $$$${the}\:{slower}\:{the}\:{pendulum}\:{or}\:{the}…
Question Number 17233 by sushmitak last updated on 02/Jul/17 $$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out}\:\mathrm{if} \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{exam}\:\mathrm{questions}. \\ $$$$\mathrm{calculators}\:\mathrm{not}\:\mathrm{allowed}. \\ $$ Commented…
Question Number 148301 by mathdanisur last updated on 26/Jul/21 $${Solve}\:{for}\:{equation}: \\ $$$$\mathrm{4}{sin}^{\mathrm{2}} \left({x}\right)\:+\:{sin}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2} \\ $$ Answered by mr W last updated on 27/Jul/21 $$\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{2} \\…
Question Number 148300 by mathdanisur last updated on 26/Jul/21 $${lg}^{\mathrm{2}} \left(\mathrm{10}{x}\right)\:+\:{lg}\left(\mathrm{10}{x}\right)\:=\:\mathrm{6}\:-\:{lg}\left({x}\right) \\ $$$${x}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\mathrm{lg}^{\mathrm{2}} \left(\mathrm{10}{x}\right)+\mathrm{lg}\left(\mathrm{10}{x}\right)\:=\:\mathrm{6}−\mathrm{lg}\left({x}\right) \\…
Question Number 82767 by jagoll last updated on 24/Feb/20 $$\mathrm{If}\:\frac{\mathrm{sin}\:\left(\mathrm{A}+\theta\right)}{\mathrm{sin}\:\left({B}+\theta\right)}\:=\:\sqrt{\frac{\mathrm{sin}\:\mathrm{2}{A}}{\mathrm{sin}\:\mathrm{2}{B}}} \\ $$$${prove}\:{that}\:\mathrm{tan}\:^{\mathrm{2}} \theta\:=\:\mathrm{tan}\:{A}.\mathrm{tan}\:{B} \\ $$ Commented by jagoll last updated on 24/Feb/20 $${thank}\:{you}\:{mr}\:{w}\:{and}\:{john} \\ $$…
Question Number 148303 by mathmax by abdo last updated on 26/Jul/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)} \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$ Answered by Olaf_Thorendsen last updated on 27/Jul/21 $${f}\left({x}\right)\:=\:\frac{\mathrm{cos}\left(\mathrm{2}{x}\right)}{\mathrm{sin}{x}} \\…
Question Number 148302 by mathmax by abdo last updated on 26/Jul/21 $$\mathrm{calculate}\:\:\int_{\mid\mathrm{z}\mid=\mathrm{3}} \:\:\:\frac{\mathrm{cos}\left(\mathrm{2iz}\right)}{\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$ Answered by Olaf_Thorendsen last updated on 27/Jul/21 $${f}\left({z}\right)\:=\:\frac{\mathrm{cos}\left(\mathrm{2}{iz}\right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }…
Question Number 82761 by TANMAY PANACEA last updated on 24/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({cosx}\right)^{\frac{\mathrm{1}}{{m}}} −\left({cosx}\right)^{\frac{\mathrm{1}}{{n}}} }{{x}^{\mathrm{2}} }\:\:\left[{where}\:{m}\:{and}\:{n}\:{integer}\right] \\ $$ Commented by mr W last updated on 24/Feb/20…
Question Number 82759 by jagoll last updated on 24/Feb/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2cos}\:\mathrm{80}^{{o}} }}}}\:=\: \\ $$ Answered by jagoll last updated on 24/Feb/20 Terms of Service…
Question Number 17220 by Arnab Maiti last updated on 02/Jul/17 $$\mathrm{Show}\:\mathrm{that}\:\int_{\mathrm{a}} ^{\:\mathrm{b}} {f}\left(\mathrm{kx}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{k}}\int_{\mathrm{ka}} ^{\:\mathrm{kb}} {f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Answered by ajfour last updated on 02/Jul/17 $$\mathrm{let}\:\mathrm{kx}=\mathrm{t}\:\:\:\Rightarrow\:\:\:\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{k}}…