Question-29413

Question Number 29413 by math solver last updated on 08/Feb/18

Commented by math solver last updated on 08/Feb/18

$${q}.\mathrm{3}\:? \\ $$

Answered by beh.i83417@gmail.com last updated on 09/Feb/18

x^2 −2x.cosα+cos^2 α+sin^2 α=0 y^2 −2y.cosβ+cos^2 β+sin^2 β=0 (x−cosα)^2 =i^2 sin^2 α⇒x=cosα±i.sinα (y−cosβ)^2 =i^2 sin^2 β⇒y=cosβ±i.sinβ (±ve): { ((x=cosα+i.sinα)),(((1/x)=cosα−i.sinα)) :}⇒ { ((cosα=(1/2)(x+(1/x)))),((sinα=(1/(2i))(x−(1/x)))) :} cos(α+β)=cosα.cosβ−sinα.sinβ cosα.cosβ=(1/4).(((x^2 +1)(y^2 +1))/(xy)) sinα.sinβ=((−1)/4).(((x^2 −1)(y^2 −1))/(xy)) 4cos(α+β)=((x^2 .y^2 +x^2 +y^2 +1+x^2 .y^2 −x^2 −y^2 +1)/(xy)) ⇒2cos(α+β)=((x^2 y^2 +1)/(xy))=xy+(1/(xy)) .■

$${x}^{\mathrm{2}} −\mathrm{2}{x}.{cos}\alpha+{cos}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \alpha=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}.{cos}\beta+{cos}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \beta=\mathrm{0} \\ $$$$\left({x}−{cos}\alpha\right)^{\mathrm{2}} ={i}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha\Rightarrow{x}={cos}\alpha\pm{i}.{sin}\alpha \\ $$$$\left({y}−{cos}\beta\right)^{\mathrm{2}} ={i}^{\mathrm{2}} {sin}^{\mathrm{2}} \beta\Rightarrow{y}={cos}\beta\pm{i}.{sin}\beta \\ $$$$\left(\pm{ve}\right): \\ $$$$\begin{cases}{{x}={cos}\alpha+{i}.{sin}\alpha}\\{\frac{\mathrm{1}}{{x}}={cos}\alpha−{i}.{sin}\alpha}\end{cases}\Rightarrow\begin{cases}{{cos}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right)}\\{{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}{i}}\left({x}−\frac{\mathrm{1}}{{x}}\right)}\end{cases} \\ $$$${cos}\left(\alpha+\beta\right)={cos}\alpha.{cos}\beta−{sin}\alpha.{sin}\beta \\ $$$${cos}\alpha.{cos}\beta=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{{xy}} \\ $$$${sin}\alpha.{sin}\beta=\frac{−\mathrm{1}}{\mathrm{4}}.\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}{{xy}} \\ $$$$\mathrm{4}{cos}\left(\alpha+\beta\right)=\frac{{x}^{\mathrm{2}} .{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} .{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{1}}{{xy}} \\ $$$$\Rightarrow\mathrm{2}{cos}\left(\alpha+\beta\right)=\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}{{xy}}={xy}+\frac{\mathrm{1}}{{xy}}\:.\blacksquare \\ $$

Commented by math solver last updated on 09/Feb/18