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Question Number 6762 by 314159 last updated on 24/Jul/16
Commented by Yozzii last updated on 24/Jul/16
x_1 ^2 +x_2 ^2 +2x_1 x_2 cosθ=1, y_1 ^2 +y_2 ^2 +2y_1 y_2 cosθ=1, x_1 y_1 +x_2 y_2 +(x_1 y_2 +x_2 y_1 )cosθ=0. Prove that x_1 ^2 +y_1 ^2 =cosec^2 θ.
$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} {cos}\theta=\mathrm{1}, \\ $$$${y}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{y}_{\mathrm{1}} {y}_{\mathrm{2}} {cos}\theta=\mathrm{1}, \\ $$$${x}_{\mathrm{1}} {y}_{\mathrm{1}} +{x}_{\mathrm{2}} {y}_{\mathrm{2}} +\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} +{x}_{\mathrm{2}} {y}_{\mathrm{1}} \right){cos}\theta=\mathrm{0}. \\ $$$$ \\ $$$${Prove}\:{that}\:{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} ={cosec}^{\mathrm{2}} \theta. \\ $$$$ \\ $$
Commented by Yozzii last updated on 24/Jul/16
x_1 ^2 +x_2 ^2 +2x_1 x_2 cosθ=1......................(1) y_1 ^2 +y_2 ^2 +2y_1 y_2 cosθ=1.......................(2) x_1 x_2 +y_1 y_2 +(x_1 y_2 +y_1 x_2 )cosθ=0......(3) or x_1 x_2 +y_1 y_2 =−(x_1 y_2 +y_1 x_2 )cosθ for (3). We start by finding x_2 in terms of x_1 in equation (1). (x_2 +x_1 cosθ)^2 +x_1 ^2 =x_2 ^2 +2x_1 x_2 cosθ+x_1 ^2 +x_1 ^2 cos^2 θ (x_2 +x_1 cosθ)^2 +x_1 ^2 =1+x_1 ^2 cos^2 θ (x_2 +x_1 cosθ)^2 =1−x_1 ^2 sin^2 θ x_2 =−x_1 cosθ±(√(1−x_1 ^2 sin^2 θ)) Similarly, y_2 in terms of y_1 for equation (2) is given by y_2 =−y_1 cosθ±(√(1−y_1 ^2 sin^2 θ)). −−−−−−−−−−−−−−−−−−−−−−−−−−− Working at the second form of equation (3) term by term with x_2 and y_2 substituted leads to the following results. (y_1 x_2 +x_1 y_2 )cosθ=−2x_1 y_1 cos^2 θ±(y_1 (√(1−x_1 ^2 sin^2 θ))+x_1 (√(1−y_1 ^2 sin^2 θ)))cosθ −(y_1 x_2 +x_1 y_2 )cosθ=2x_1 y_1 cos^2 θ∓k {Let k=(y_1 (√(1−x_1 ^2 sin^2 θ))+x_1 (√(1−y_1 ^2 sin^2 θ)))cosθ} and x_2 y_2 =x_1 y_1 cos^2 θ∓{x_1 (√(1−y_1 ^2 sin^2 θ))+y_1 (√(1−x_1 ^2 sin^2 θ))}cosθ+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) x_2 y_2 =x_1 y_1 +x_1 y_1 cos^2 θ∓k+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) ∴ second form of (3) becomes, after cancellation of ∓k from both sides, x_1 y_1 +x_1 y_1 cos^2 θ+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=2x_1 y_1 cos^2 θ x_1 y_1 (1+cos^2 θ−2cos^2 θ)+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=0 x_1 y_1 (1−cos^2 θ)+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=0 x_1 y_1 sin^2 θ=−(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) ⇒(1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)=x_1 ^2 y_1 ^2 sin^4 θ 1−x_1 ^2 sin^2 θ−y_1 ^2 sin^2 θ+x_1 ^2 y_1 ^2 sin^4 θ=x_1 ^2 y_1 ^2 sin^4 θ 1−(x_1 ^2 +y_1 ^2 )sin^2 θ=0 sin^2 θ=(1/(x_1 ^2 +y_1 ^2 ))⇒x_1 ^2 +y_1 ^2 =cosec^2 θ This result assumes that in x_2 and y_2 that both ± signs take the same signs. −−−−−−−−−−−−−−−−−−−−−−−−−−−− Suppose that ± takes different signs between x_2 and y_2 ; i.e x_2 =−x_1 cosθ+(√(1−x_1 ^2 sin^2 θ)) y_2 =−y_1 cosθ−(√(1−y_1 ^2 sin^2 θ)). You can show still that cosec^2 θ=x_1 ^2 +y_1 ^2 . ////
$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} {cos}\theta=\mathrm{1}………………….\left(\mathrm{1}\right) \\ $$$${y}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{y}_{\mathrm{1}} {y}_{\mathrm{2}} {cos}\theta=\mathrm{1}…………………..\left(\mathrm{2}\right) \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} +{y}_{\mathrm{1}} {x}_{\mathrm{2}} \right){cos}\theta=\mathrm{0}……\left(\mathrm{3}\right) \\ $$$${or}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} =−\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} +{y}_{\mathrm{1}} {x}_{\mathrm{2}} \right){cos}\theta\:\:{for}\:\left(\mathrm{3}\right). \\ $$$$ \\ $$$${We}\:{start}\:{by}\:{finding}\:{x}_{\mathrm{2}} \:{in}\:{terms}\:{of}\:{x}_{\mathrm{1}} \:{in}\:{equation}\:\left(\mathrm{1}\right). \\ $$$$\left({x}_{\mathrm{2}} +{x}_{\mathrm{1}} {cos}\theta\right)^{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{2}} ={x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} {cos}\theta+{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{2}} {cos}^{\mathrm{2}} \theta \\ $$$$\left({x}_{\mathrm{2}} +{x}_{\mathrm{1}} {cos}\theta\right)^{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}} {cos}^{\mathrm{2}} \theta \\ $$$$\left({x}_{\mathrm{2}} +{x}_{\mathrm{1}} {cos}\theta\right)^{\mathrm{2}} =\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$$${x}_{\mathrm{2}} =−{x}_{\mathrm{1}} {cos}\theta\pm\sqrt{\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta} \\ $$$$ \\ $$$${Similarly},\:{y}_{\mathrm{2}} \:{in}\:{terms}\:{of}\:{y}_{\mathrm{1}} \:{for}\:{equation}\:\left(\mathrm{2}\right)\:{is}\:{given}\:{by} \\ $$$${y}_{\mathrm{2}} =−{y}_{\mathrm{1}} {cos}\theta\pm\sqrt{\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Working}\:{at}\:{the}\:{second}\:{form}\:{of}\:{equation}\:\left(\mathrm{3}\right)\:{term}\:{by}\:{term} \\ $$$${with}\:{x}_{\mathrm{2}} \:{and}\:{y}_{\mathrm{2}} \:{substituted}\:{leads}\:{to}\:{the}\:{following}\:{results}. \\ $$$$\left({y}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} {y}_{\mathrm{2}} \right){cos}\theta=−\mathrm{2}{x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta\pm\left({y}_{\mathrm{1}} \sqrt{\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}+{x}_{\mathrm{1}} \sqrt{\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\right){cos}\theta \\ $$$$−\left({y}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} {y}_{\mathrm{2}} \right){cos}\theta=\mathrm{2}{x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta\mp{k}\:\:\:\left\{{Let}\:{k}=\left({y}_{\mathrm{1}} \sqrt{\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}+{x}_{\mathrm{1}} \sqrt{\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\right){cos}\theta\right\} \\ $$$${and} \\ $$$${x}_{\mathrm{2}} {y}_{\mathrm{2}} ={x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta\mp\left\{{x}_{\mathrm{1}} \sqrt{\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}+{y}_{\mathrm{1}} \sqrt{\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\right\}{cos}\theta+\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)} \\ $$$${x}_{\mathrm{2}} {y}_{\mathrm{2}} ={x}_{\mathrm{1}} {y}_{\mathrm{1}} +{x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta\mp{k}+\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)} \\ $$$$ \\ $$$$\therefore\:{second}\:{form}\:{of}\:\left(\mathrm{3}\right)\:{becomes},\:{after}\:{cancellation}\:{of}\:\mp{k}\:{from}\:{both}\:{sides}, \\ $$$${x}_{\mathrm{1}} {y}_{\mathrm{1}} +{x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta+\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)}=\mathrm{2}{x}_{\mathrm{1}} {y}_{\mathrm{1}} {cos}^{\mathrm{2}} \theta \\ $$$${x}_{\mathrm{1}} {y}_{\mathrm{1}} \left(\mathrm{1}+{cos}^{\mathrm{2}} \theta−\mathrm{2}{cos}^{\mathrm{2}} \theta\right)+\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} {y}_{\mathrm{1}} \left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)+\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} {y}_{\mathrm{1}} {sin}^{\mathrm{2}} \theta=−\sqrt{\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)} \\ $$$$\Rightarrow\left(\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)\left(\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right)={x}_{\mathrm{1}} ^{\mathrm{2}} {y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{4}} \theta \\ $$$$\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta+{x}_{\mathrm{1}} ^{\mathrm{2}} {y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{4}} \theta={x}_{\mathrm{1}} ^{\mathrm{2}} {y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{4}} \theta \\ $$$$\mathrm{1}−\left({x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} \right){sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$${sin}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} }\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} ={cosec}^{\mathrm{2}} \theta\:\:\:\:\:\:\:\:\: \\ $$$${This}\:{result}\:{assumes}\:{that}\:{in}\:{x}_{\mathrm{2}} \:{and}\:{y}_{\mathrm{2}} \\ $$$${that}\:{both}\:\pm\:{signs}\:{take}\:{the}\:{same}\:{signs}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Suppose}\:{that}\:\pm\:{takes}\:{different}\:{signs} \\ $$$${between}\:{x}_{\mathrm{2}} \:{and}\:{y}_{\mathrm{2}} ;\:{i}.{e} \\ $$$${x}_{\mathrm{2}} =−{x}_{\mathrm{1}} {cos}\theta+\sqrt{\mathrm{1}−{x}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta} \\ $$$${y}_{\mathrm{2}} =−{y}_{\mathrm{1}} {cos}\theta−\sqrt{\mathrm{1}−{y}_{\mathrm{1}} ^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}. \\ $$$${You}\:{can}\:{show}\:{still}\:{that}\:{cosec}^{\mathrm{2}} \theta={x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} .\:\:\:\://// \\ $$
Answered by Yozzii last updated on 24/Jul/16
Check comments for a solution I shared.
$${Check}\:{comments}\:{for}\:{a}\:{solution}\:{I}\:{shared}. \\ $$

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