Question Number 184932 by mathlove last updated on 14/Jan/23 Answered by qaz last updated on 14/Jan/23 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}=\Im\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{i}\frac{{n}\pi}{\mathrm{3}}} }{{n}}=−\Im{ln}\left(\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right) \\ $$$$=−\Im{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)=−\Im{lne}^{−{i}\mathrm{arctan}\:\sqrt{\mathrm{3}}}…
Question Number 119373 by bobhans last updated on 24/Oct/20 $${Find}\:{all}\:{sum}\:{of}\:{all}\:{x}\:{interval} \\ $$$$\left[\:\mathrm{0},\:\mathrm{2}\pi\:\right]\:{such}\:{that}\:\mathrm{3cot}\:^{\mathrm{2}} {x}\:+\:\mathrm{8}\:\mathrm{cot}\:{x}\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$ Commented by benjo_mathlover last updated on 24/Oct/20 $${consider}\:{the}\:{quadratic}\:{equation} \\ $$$$\mathrm{3}{q}^{\mathrm{2}}…
Question Number 119372 by bobhans last updated on 24/Oct/20 $${Determine}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\:,\:{over}\:{all}\:\alpha,\beta\:\neq\:\frac{{k}\pi}{\mathrm{2}} \\ $$$${and}\:{k}\in\mathbb{Z} \\ $$ Answered by 1549442205PVT last updated…
Question Number 119295 by bobhans last updated on 23/Oct/20 $$\:{Let}\:{a}\:{and}\:{b}\:\:{non}\:{negative}\:{real}\:{numbers}\: \\ $$$${If}\:\mathrm{sin}\:{x}+{a}\mathrm{cos}\:{x}={b}\:,\:{express}\: \\ $$$$\mid\:{a}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$$$ \\ $$ Answered by bemath last updated on 23/Oct/20…
Question Number 119266 by benjo_mathlover last updated on 23/Oct/20 $$\:{solve}\:\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${where}\:{x}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$ Commented by MJS_new last updated on 23/Oct/20 $$\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{sin}\:{x}\:−\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\mathrm{cos}\:{x}\:=\mathrm{4}\sqrt{\mathrm{2}}\mathrm{sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\left[\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{to}\:\mathrm{type}\:\mathrm{this}\:\mathrm{transformation}\right] \\…
Question Number 119246 by 1549442205PVT last updated on 23/Oct/20 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\:\mathrm{inequalities}\:\mathrm{hold} \\ $$$$\:\:\mathrm{true}\:\forall\mathrm{x}\in\mathrm{R} \\ $$$$\left.\mathrm{a}\right)\mathrm{cos}\left(\mathrm{cosx}\right)>\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{cos}\left(\mathrm{sinx}\right)>\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$ Answered by Olaf last updated on 23/Oct/20…
Question Number 119229 by benjo_mathlover last updated on 23/Oct/20 $$\:\:\:\mathrm{sin}\:\mathrm{3}{A}+\mathrm{cos}\:\mathrm{3}{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by bemath last updated on 23/Oct/20 $${let}\:\mathrm{3}{A}\:=\:{x}\:;\:\mathrm{cos}\:{x}\:=\:{X}\:{and}\:\mathrm{sin}\:{x}\:=\:{Y} \\ $$$$\Rightarrow{X}+{Y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:{X}^{\mathrm{2}} +\mathrm{2}{XY}\:+{Y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}} \\…
Question Number 119220 by benjo_mathlover last updated on 23/Oct/20 $$\:\:\:\:\:\mathrm{2cos}\:^{\mathrm{2}} {x}\:−\mathrm{25}\:\mathrm{cot}\:{x}\:=\:\mathrm{0}\: \\ $$$$\:\:\:\:\:{x}\:=? \\ $$ Answered by MJS_new last updated on 23/Oct/20 $$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \\…
Question Number 119109 by bagjagunawan last updated on 22/Oct/20 Answered by 1549442205PVT last updated on 22/Oct/20 $$\mathrm{P}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{9}\pi}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{27}\pi}{\mathrm{20}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos9}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos81}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos27}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos9}+\mathrm{cos81}}{\mathrm{2}}+\mathrm{cos9cos81}\right) \\ $$$$×\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos27}−\mathrm{cos63}}{\mathrm{2}}−\mathrm{cos27cos63}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{cos45cos36}+\frac{\mathrm{cos90}+\mathrm{cos72}}{\mathrm{2}}\right)…
Question Number 119015 by benjo_mathlover last updated on 21/Oct/20 $${x}\:=\:\frac{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{45}°}{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{45}°} \\ $$$${x}\:=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 21/Oct/20 $${tan}\left(\mathrm{1}°+\frac{\mathrm{45}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)={tan}\mathrm{23}° \\ $$…