Question Number 32164 by jarjum last updated on 20/Mar/18 $${Determine}\:{wether}\:{the}\:{relation}\:{on}\:{the} \\ $$$${set}\:{R}\:{of}\:{all}\:{real}\:{number}\:{as} \\ $$$${R}=\left\{\left({a},{b}\right):{a},{b}\in{R}\:{and}\:{a}−{b}\:+\sqrt{\mathrm{3}}\:\in\:{s}\:\right. \\ $$$$\left.{where}\:{s}\:{is}\:{the}\:{set}\:{of}\:{all}\:{irrational}\:{no}\right) \\ $$$${is}\:{reflexive},\:{symmetric}\:{and}\:{transitive}? \\ $$ Terms of Service Privacy Policy…
Question Number 32161 by jarjum last updated on 20/Mar/18 $${Let}\:{a}\:{function}\:{F}\::{R}\rightarrow{R}\:{be}\:{defined}\:{by} \\ $$$${f}\left({x}\right)=\mathrm{1}+{ax},\alpha\neq\:\mathrm{0}\:{for}\:{all}\:{X}\:\in\:{R}.\:{Show} \\ $$$${that}\:{f}\:{is}\:{invertible}\:{and}\:{find}\:{its}\:{inverse} \\ $$$${function}.{Also}\:{find}\:{the}\:{value}\:\left({s}\right)\:{of}\:\alpha \\ $$$${if}\:{inverse}\:{of}\:{f}\:{is}\:{itself} \\ $$ Answered by mrW2 last updated…
Question Number 32156 by jarjum last updated on 20/Mar/18 $${evaluate}\int\frac{\sqrt{{cos}\:\mathrm{2}{x}}}{{sin}\:{x}}{dx} \\ $$ Commented by abdo imad last updated on 20/Mar/18 $${I}\:=\:\int\:\sqrt{\frac{{cos}\left(\mathrm{2}{x}\right)}{{sin}^{\mathrm{2}} {x}}}\:{dx}\:=\int\:\sqrt{\frac{{cos}\left(\mathrm{2}{x}\right)}{\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}}\:{dx} \\ $$$$=\int\:\:\sqrt{\:\:\:\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}}\:{dx}\:\:{ch}.\:{tanx}={t}\:{give} \\…
Question Number 97680 by bemath last updated on 09/Jun/20 Commented by bobhans last updated on 09/Jun/20 $$\mathrm{cos}\:\mathrm{6A}−\mathrm{cos}\:\mathrm{2A}=−\mathrm{2sin}\:\mathrm{4Asin}\:\mathrm{2A} \\ $$$$\Leftrightarrow−\mathrm{2sin}\:\mathrm{4Asin}\:\mathrm{2A}−\mathrm{2cos}\:\mathrm{4A}+\mathrm{2}= \\ $$$$\Leftrightarrow\:−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{2A}\:\mathrm{cos}\:\mathrm{2A}−\mathrm{2}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2A}\right)+\mathrm{2} \\ $$$$\Leftrightarrow−\mathrm{4sin}\:^{\mathrm{2}}…
Question Number 97546 by bobhans last updated on 08/Jun/20 Commented by bobhans last updated on 08/Jun/20 can anyone translate to english? Commented by bemath last updated on 08/Jun/20 $$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:…
Question Number 163020 by amin96 last updated on 03/Jan/22 $$\boldsymbol{\mathrm{cos}}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{sin}}^{\mathrm{2}\boldsymbol{\mathrm{n}}} \left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{4}^{\boldsymbol{\mathrm{n}}} +\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$ Commented by amin96 last updated on 03/Jan/22…
Question Number 97346 by bobhans last updated on 07/Jun/20 Answered by john santu last updated on 07/Jun/20 $$\mathrm{1}+\mathrm{3}\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}}\:\sqrt[{\mathrm{3}\:\:}]{\left(\mathrm{sin}\:\mathrm{xcos}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\mathrm{3}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{2}}\:\sqrt[{\mathrm{3}\:\:}]{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2x}}=\:\mathrm{1} \\ $$$$\mathrm{54}\:\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{2x}\right)\:=\:\mathrm{1}…
Question Number 31808 by Abadon last updated on 15/Mar/18 $$\mathrm{The}\:\mathrm{numeric}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{is}: \\ $$$$ \\ $$$$\frac{\mathrm{Sec}\:\mathrm{1320}°}{\mathrm{2}}\:−\:\mathrm{2}\:\centerdot\:\mathrm{cos}\:\left(\frac{\mathrm{53}\pi}{\mathrm{3}}\right)\:+\:\left(\mathrm{tg}\:\mathrm{2220}°\right)^{\mathrm{2}} \\ $$ Commented by Tinkutara last updated on 15/Mar/18 1 Commented…
Question Number 162872 by amin96 last updated on 01/Jan/22 Commented by amin96 last updated on 01/Jan/22 $$\boldsymbol{{PROVE}}\:\:\:\:\:\boldsymbol{{THAT}} \\ $$ Answered by Ar Brandon last updated…
Question Number 31745 by pieroo last updated on 13/Mar/18 $$\mathrm{pls},\:\mathrm{why}\:\mathrm{is}\:\mathrm{cos}^{\mathrm{4}} −\mathrm{sin}^{\mathrm{4}\:} \:=\:\mathrm{cos}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} ? \\ $$ Answered by Tinkutara last updated on 13/Mar/18 $$\mathrm{cos}^{\mathrm{4}} \:\theta−\mathrm{sin}^{\mathrm{4}}…