Question Number 94241 by Ar Brandon last updated on 17/May/20 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\mathrm{arctan}\left(\mathrm{x}\right)+\mathrm{2arctan}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$ Commented by mathmax by abdo last updated on 17/May/20…
Question Number 28648 by tawa tawa last updated on 28/Jan/18 $$\mathrm{If}\:\:\theta\:\:=\:\:\mathrm{log}_{\mathrm{e}} \left[\mathrm{tanh}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right]\:,\:\:\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\mathrm{3tanh}\left(\mathrm{2}\theta\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28624 by Joel578 last updated on 28/Jan/18 Commented by Joel578 last updated on 28/Jan/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression} \\ $$ Commented by Tinkutara last updated on…
Question Number 28614 by abdo imad last updated on 27/Jan/18 $${find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}}. \\ $$$$ \\ $$ Commented by abdo imad last updated on 30/Jan/18…
Question Number 28609 by abdo imad last updated on 27/Jan/18 $${calculate}\:{cotanx}\:−\mathrm{2}{cotan}\left(\mathrm{2}{x}\right){then}\:{simlify} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }{tan}\left(\frac{\alpha}{\mathrm{2}^{{k}} }\right). \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 28608 by abdo imad last updated on 27/Jan/18 $${transform}\:{tanp}−{tanq}\:{then}\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{cos}\left({k}\theta\right){cos}\left(\left({k}+\mathrm{1}\right)\theta\right.}\:.\:\:\theta\in{R}. \\ $$ Answered by Tinkutara last updated on 28/Jan/18 $$\frac{\mathrm{1}}{\mathrm{cos}\:{k}\theta\mathrm{cos}\:\left({k}+\mathrm{1}\right)\theta}…
Question Number 28583 by tawa tawa last updated on 27/Jan/18 Commented by tawa tawa last updated on 27/Jan/18 $$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{above}:\:\:\:\:\mathrm{AB}\:\mathrm{is}\:\mathrm{a}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{10cm}.\:\mathrm{M}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{mid}\:\mathrm{point}\:\mathrm{of}\:\mathrm{AB}.\:\mathrm{and}\:\:\mathrm{NM}\:\bot\:\mathrm{AB} \\ $$ Answered by…
Question Number 159641 by tounghoungko last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{f}\left({x}\right)=\mathrm{256}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)+\mathrm{324}\:\mathrm{cosec}\:^{\mathrm{2}} \left({x}\right) \\ $$$$\:\forall{x}\in\:\mathbb{R}\: \\ $$ Commented by gsk2684 last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{f}\left({x}\right)=\mathrm{256}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)+\mathrm{324}\:\mathrm{cosec}\:^{\mathrm{2}}…
Question Number 159642 by tounghoungko last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{function}\: \\ $$$$\:\:\:{f}\left({x}\right)=\sqrt{\left(\mathrm{3sin}\:{x}−\mathrm{4cos}\:{x}−\mathrm{10}\right)\left(\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}−\mathrm{10}\right)} \\ $$ Commented by bobhans last updated on 19/Nov/21 $$\:\mathrm{g}\left(\mathrm{x}\right)=\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)−\mathrm{4cos}\:\mathrm{x}\right)\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)+\mathrm{4cos}\:\mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)^{\mathrm{2}} −\mathrm{16cos}\:^{\mathrm{2}}…
Question Number 159635 by cortano last updated on 19/Nov/21 Commented by cortano last updated on 19/Nov/21 $$\:\mathrm{tan}\:\alpha\:=\:\frac{{x}}{\mathrm{4}}\:;\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{3}{x}}{\mathrm{12}\sqrt{\mathrm{3}}}\:=\:\frac{{x}}{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\alpha\: \\ $$$$\Rightarrow\alpha+\beta+\mathrm{105}°=\mathrm{180}° \\ $$$$\Rightarrow\alpha+\beta\:=\:\mathrm{75}° \\ $$$$\Rightarrow\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:…