Question Number 149137 by mathdanisur last updated on 03/Aug/21 $${if}\:\:\:{ctg}\boldsymbol{\alpha}\:=\:−\mathrm{2}\:\:\:{and}\:\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:<\:\alpha\:<\:\mathrm{2}\pi \\ $$$${find}\:\:\:{sin}\boldsymbol{\alpha}\:=\:? \\ $$ Commented by liberty last updated on 03/Aug/21 $$\mathrm{sin}\:\alpha=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$ Terms…
Question Number 83603 by jagoll last updated on 04/Mar/20 $$\int\:\frac{\mathrm{dx}}{\mathrm{1}−\mathrm{2cos}\:\mathrm{x}} \\ $$ Commented by turbo msup by abdo last updated on 04/Mar/20 $${we}\:{use}\:{the}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$$\Rightarrow\int\:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{cosx}}\:=\int\:\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\frac{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 18066 by tawa tawa last updated on 14/Jul/17 $$\left(\mathrm{a}\right)\:\:\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}:\:\:\mathrm{y}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{3x}\:+\:\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} \:−\:\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\left(\mathrm{b}\right)\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{identity}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} \:−\:\mathrm{ax}}{\left(\mathrm{x}\:−\:\mathrm{2a}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{a}^{\mathrm{2}} \right)}\:\equiv\:\frac{\mathrm{A}}{\left(\mathrm{x}\:−\:\mathrm{2a}\right)}\:+\:\frac{\mathrm{Bx}\:+\:\mathrm{Ca}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{where}\:\:\mathrm{a}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\:\:\mathrm{hence}\:\mathrm{prove}\:\mathrm{that}.\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}}…
Question Number 18063 by Tinkutara last updated on 14/Jul/17 $$\mathrm{The}\:\mathrm{angles}\:{A},\:{B},\:{C}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{satisfy}\:\mathrm{4cos}{A}\mathrm{cos}{B}\:+\:\mathrm{sin2}{A}\:+\:\mathrm{sin2}{B}\:+ \\ $$$$\mathrm{sin2}{C}\:=\:\mathrm{4}.\:\mathrm{Then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{statements}\:\mathrm{is}/\mathrm{are}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{isosceles} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{neither} \\ $$$$\mathrm{isosceles}\:\mathrm{nor}\:\mathrm{right}\:\mathrm{angled} \\…
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Question Number 18062 by Tinkutara last updated on 14/Jul/17 $$\mathrm{If}\:\mathrm{0}\:<\:\alpha,\:\beta\:<\:\pi\:\mathrm{and}\:\mathrm{they}\:\mathrm{satisfy} \\ $$$$\mathrm{cos}\:\alpha\:+\:\mathrm{cos}\:\beta\:−\:\mathrm{cos}\:\left(\alpha\:+\:\beta\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\alpha\:=\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:+\:\beta\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:\alpha\:=\:\mathrm{2}\beta \\ $$$$\left(\mathrm{4}\right)\:\beta\:=\:\mathrm{2}\alpha \\ $$ Answered by Tinkutara…
Question Number 83597 by dennis kenneth last updated on 04/Mar/20 $$\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\right){dx} \\ $$ Commented by Jidda28 last updated on 04/Mar/20 $$\left.=\:\:{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}}…
Question Number 149129 by Samimsultani last updated on 03/Aug/21 Commented by ajfour last updated on 03/Aug/21 $${isn}'{t}\:\:\boldsymbol{\div}\:\:{same}\:{as}\:/\: \\ $$ Answered by puissant last updated on…
Question Number 149128 by liberty last updated on 03/Aug/21 $$ \\ $$does anyone in this group have monbusho test questions? Terms of Service Privacy Policy…
Question Number 83590 by Tony Lin last updated on 04/Mar/20 $${transform}\:{the}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:{to} \\ $$$${the}\:{polar}\:{equation}\:{r}=\:\frac{{a}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{\mathrm{1}+{ecos}\theta} \\ $$$${a}:\:{semimajor}\:{axis} \\ $$$${e}:\:{eccentricity} \\ $$ Commented…