Question Number 118078 by bemath last updated on 15/Oct/20 $$\mathrm{3}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} − \\ $$$$\mathrm{3}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right)\:=\:?\: \\ $$ Answered by bobhans last updated on 15/Oct/20…
Question Number 118070 by andilizhaa last updated on 15/Oct/20 Commented by bemath last updated on 15/Oct/20 $$\mathrm{see}\:\mathrm{qn}\:\mathrm{118064} \\ $$ Commented by andilizhaa last updated on…
Question Number 118065 by andilizhaa last updated on 15/Oct/20 Answered by bobhans last updated on 15/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{cot}\:\mathrm{x} \\ $$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\mathrm{cosec}\:^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\mid_{\frac{\pi}{\mathrm{6}}} \:=\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)}\:=\:\mathrm{4} \\…
Question Number 118064 by andilizhaa last updated on 15/Oct/20 Answered by bobhans last updated on 15/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}}]{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\right)^{\mathrm{4}} } \\ $$$$\mathrm{by}\:\mathrm{chain}\:\mathrm{rule} \\ $$$$\mathrm{letting}\:\mathrm{u}\:=\:\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\:\rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{20x} \\…
Question Number 118062 by andilizhaa last updated on 15/Oct/20 Commented by bobhans last updated on 15/Oct/20 $$\mathrm{y}'=\mathrm{3}×\mathrm{6}×\mathrm{sin}\:^{\mathrm{5}} \left(\mathrm{3x}−\pi\right)×\mathrm{cos}\:\left(\mathrm{3x}−\pi\right) \\ $$$$\mathrm{y}'=\mathrm{9sin}\:^{\mathrm{4}} \left(\mathrm{3x}−\pi\right)\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{2}\pi\right) \\ $$ Commented by…
Question Number 117960 by andilizhaa last updated on 14/Oct/20 Answered by mathmax by abdo last updated on 14/Oct/20 $$\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{derivative}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5x}^{\mathrm{4}} \:+\mathrm{x}\sqrt{\mathrm{x}}+\frac{\mathrm{6}}{\:\sqrt{\mathrm{x}}}\:−\mathrm{sinx}−\mathrm{2cosx}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{5x}^{\mathrm{4}} \:+\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 117943 by bemath last updated on 14/Oct/20 $$\mathrm{solve}\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{112}°}{\mathrm{16sin}\:\mathrm{7}°}\:−\mathrm{cos}\:\mathrm{7}°.\mathrm{cos}\:\mathrm{14}°.\mathrm{cos}\:\mathrm{28}°.\mathrm{cos}\:\mathrm{56}°\:=? \\ $$ Answered by TANMAY PANACEA last updated on 14/Oct/20 $${p}={cos}\mathrm{7}{cos}\mathrm{14}{cos}\mathrm{28}{cos}\mathrm{56} \\ $$$$\mathrm{2}{psin}\mathrm{7}={sin}\mathrm{14}{cos}\mathrm{14}{cos}\mathrm{28}{cos}\mathrm{56} \\ $$$$\mathrm{2}^{\mathrm{2}}…
Question Number 117938 by bemath last updated on 14/Oct/20 $$\left(\mathrm{1}\right)\:\frac{\mid{a}^{\mathrm{2}} −\mathrm{16}\mid}{\mathrm{4}−{a}}\:−\:\frac{\mid{a}^{\mathrm{2}} −\mathrm{9}\mid}{\mathrm{3}+{a}}\:−\:\frac{\mid\mathrm{4}−{a}^{\mathrm{2}} \mid}{\mathrm{2}−{a}}\:=? \\ $$$$\left(\mathrm{2}\right)\left(\frac{\mid\mathrm{x}\mid\:+\mathrm{x}\:}{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{14x}}{\mathrm{x}−\mathrm{1}}\:+\:\mathrm{12}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{2}^{\mathrm{2x}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}−\mathrm{6sin}\:^{\mathrm{2}} \mathrm{x}\:\right)\:=\:\mathrm{x} \\ $$$$\mathrm{where}\:\frac{\mathrm{5}\pi}{\mathrm{2}}\:\leqslant\mathrm{x}\leqslant\mathrm{4}\pi \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{2cos}\:^{\mathrm{2}}…
Question Number 117926 by bemath last updated on 14/Oct/20 $$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+\:\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2x}}\right)\:=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} +\left(\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} =\frac{\mathrm{5}\pi}{\mathrm{8}} \\ $$$$ \\ $$ Answered…
Question Number 117852 by bemath last updated on 14/Oct/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{24}}+\mathrm{tan}\:\frac{\mathrm{5}\pi}{\mathrm{24}}\right).\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:+\:\mathrm{2}\:. \\ $$$$\left(\mathrm{2}\right)\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5x}}}\:.\left(\mathrm{5}\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{20x}}\:\right)^{\mathrm{0}.\mathrm{5}} .\:\mathrm{2}^{−\mathrm{1}} \:=\:? \\ $$ Answered by bobhans last updated on 14/Oct/20…