Question Number 64311 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${letA}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'=\begin{bmatrix}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{cos}\frac{\pi}{\mathrm{4}}\:\:−{sin}\frac{\pi}{\mathrm{4}}}\\{{sin}\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:{cos}\frac{\pi}{\mathrm{4}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:=\begin{bmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\:\:\frac{\mathrm{9}}{\mathrm{2}}}\end{bmatrix}…
Question Number 64309 by aseer imad last updated on 16/Jul/19 Commented by Tony Lin last updated on 17/Jul/19 $${let}\:{A}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}{i} \\ $$$$\Rightarrow{D}=\left(\mathrm{1},\mathrm{3}\right)+\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$$${B}'\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{4}+\mathrm{5}{i}\right)\left({cos}\frac{\pi}{\mathrm{4}}−{isin}\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}…
Question Number 129436 by mr W last updated on 07/Feb/21 Commented by mr W last updated on 07/Feb/21 $${see}\:{Q}\mathrm{129244} \\ $$ Commented by ajfour last…
Question Number 129404 by bemath last updated on 15/Jan/21 $$\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}.\mathrm{cot}\:\left(\frac{\theta}{\mathrm{2}}\right) \\ $$$$\mathrm{prove}. \\ $$ Answered by MJS_new last updated on 15/Jan/21 $$\theta=\mathrm{2}\alpha \\ $$$$\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha\:−\mathrm{i}\:\mathrm{sin}\:\mathrm{2}\alpha\right)^{−\mathrm{1}} =…
Question Number 129243 by Engr_Jidda last updated on 14/Jan/21 $$ \\ $$ Answered by bemath last updated on 14/Jan/21 $$\:\:\:\:\:\:\: \\ $$ Commented by Engr_Jidda…
Question Number 63618 by kaivan.ahmadi last updated on 06/Jul/19 Commented by kaivan.ahmadi last updated on 06/Jul/19 $${i}\:{dont}\:{think}\: \\ $$$${if}\:{F}'{P}−{FP}=\mathrm{0}<{a}\:{then}\:{P}\:{is}\:{not}\:{inside} \\ $$ Commented by kaivan.ahmadi last…
Question Number 63336 by ajfour last updated on 02/Jul/19 Commented by ajfour last updated on 02/Jul/19 $${Find}\:{the}\:{equation}\:{of}\:{the}\:{parabola} \\ $$$${in}\:{the}\:{form}:\:\:\:{y}={Ax}^{\mathrm{2}} +{C}\:. \\ $$ Answered by mr…
Question Number 63256 by ajfour last updated on 01/Jul/19 Commented by ajfour last updated on 01/Jul/19 $${Find}\:{r}={OB},\:{such}\:{that}\:\bigtriangleup{ABC} \\ $$$${has}\:{maximum}\:{area}. \\ $$ Commented by ajfour last…
Question Number 63218 by ajfour last updated on 30/Jun/19 $${Q}.\mathrm{63108}\:\:\:\left({A}\:{check}\right) \\ $$$${eq}.\:{of}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${Inscribed}\:{equilateral}\:\bigtriangleup{ABC} \\ $$$${of}\:{side}\:\boldsymbol{{s}}=\frac{\mathrm{16}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{365}}} \\ $$$${Do}\:{these}\:{points}\:{satisfy}\:{for} \\ $$$${A},\:{B},\:{C}\:? \\…
Question Number 63178 by ajfour last updated on 30/Jun/19 Commented by Prithwish sen last updated on 30/Jun/19 $$\mathrm{Sir},\:\mathrm{whether}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\left[\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right),\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \right]\:\mathrm{or}\:\mathrm{not}\:? \\ $$ Answered by mr W…