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Question Number 39666 by ajfour last updated on 09/Jul/18

Commented by ajfour last updated on 09/Jul/18

In terms of a and 𝛉 , find x, y, 𝛂, 𝛃, and r . (y= AD)

$${In}\:{terms}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{\theta}\:,\:{find}\: \\ $$$$\:\boldsymbol{{x}},\:\boldsymbol{{y}},\:\boldsymbol{\alpha},\:\boldsymbol{\beta},\:{and}\:\boldsymbol{{r}}\:.\:\:\:\left({y}=\:{AD}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

4θ=α+β sinα=(1/r) sinβ=(a/r) cos(α+β)=((2r^2 −(a+1)^2 )/(2r^2 )) 2r^2 cos(4θ)−2r^2 =−(a+1)^2 2r^2 (1−cos4θ)=(a+1)^2 2r^2 ×2sin^2 2θ=(a+1)^2 4r^2 sin^2 2θ=(a+1)^2 2rsin2θ=(a+1) r=((a+1)/(2sin2θ)) sinα=(1/r)=((2sin2θ)/(a+1)) α=sin^(−1) (((2sin2θ)/(a+1))) sinβ=(a/r)=(a/(a+1))×2sin2(θ) β.=sin^(−1) ((a/(a+1)).2sin(2θ)} (x/y)=(1/a) so(x/1)=(y/a)k x=k and y=ak cos2θ=((x^2 +y^2 −(a+1)^2 )/(2xy)) cos2θ=((k^2 +a^2 k^2 −(a+1)^2 )/(2ak^2 )) k^2 +a^2 k^2 −2ak^2 cos2θ=(a+1)^2 k^2 (1+a^2 −2acos2θ)=(a+1)^2 k=((a+1)/(√(1+a^2 −2acos2θ))) x=((a+1)/(√(1+a^2 −2acos2θ))) y=a×((a+1)/(√(1+a^2 −2acos2θ)))

$$\mathrm{4}\theta=\alpha+\beta \\ $$$${sin}\alpha=\frac{\mathrm{1}}{{r}}\:\:\:{sin}\beta=\frac{{a}}{{r}} \\ $$$${cos}\left(\alpha+\beta\right)=\frac{\mathrm{2}{r}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} } \\ $$$$\mathrm{2}{r}^{\mathrm{2}} {cos}\left(\mathrm{4}\theta\right)−\mathrm{2}{r}^{\mathrm{2}} =−\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}−{cos}\mathrm{4}\theta\right)=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} ×\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \mathrm{2}\theta=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{rsin}\mathrm{2}\theta=\left({a}+\mathrm{1}\right) \\ $$$${r}=\frac{{a}+\mathrm{1}}{\mathrm{2}{sin}\mathrm{2}\theta} \\ $$$${sin}\alpha=\frac{\mathrm{1}}{{r}}=\frac{\mathrm{2}{sin}\mathrm{2}\theta}{{a}+\mathrm{1}} \\ $$$$\alpha={sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{sin}\mathrm{2}\theta}{{a}+\mathrm{1}}\right) \\ $$$${sin}\beta=\frac{{a}}{{r}}=\frac{{a}}{{a}+\mathrm{1}}×\mathrm{2}{sin}\mathrm{2}\left(\theta\right) \\ $$$$\beta.={sin}^{−\mathrm{1}} \left(\frac{{a}}{{a}+\mathrm{1}}.\mathrm{2}{sin}\left(\mathrm{2}\theta\right)\right\} \\ $$$$ \\ $$$$\frac{{x}}{{y}}=\frac{\mathrm{1}}{{a}}\:\:\:\:{so}\frac{{x}}{\mathrm{1}}=\frac{{y}}{{a}}{k}\:\:\:{x}={k}\:{and}\:{y}={ak} \\ $$$${cos}\mathrm{2}\theta=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{xy}} \\ $$$${cos}\mathrm{2}\theta=\frac{{k}^{\mathrm{2}} +{a}^{\mathrm{2}} {k}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{ak}^{\mathrm{2}} } \\ $$$${k}^{\mathrm{2}} +{a}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{2}{ak}^{\mathrm{2}} {cos}\mathrm{2}\theta=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} \left(\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{acos}\mathrm{2}\theta\right)=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${k}=\frac{{a}+\mathrm{1}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{acos}\mathrm{2}\theta}}\: \\ $$$${x}=\frac{{a}+\mathrm{1}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{acos}\mathrm{2}\theta}}\:\:\:{y}={a}×\frac{{a}+\mathrm{1}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{acos}\mathrm{2}\theta}} \\ $$$$ \\ $$

Commented by ajfour last updated on 10/Jul/18

Excellent, Tanmay Sir. Thank you.

$$\mathscr{E}{xcellent},\:{Tanmay}\:{Sir}.\:{Thank}\:{you}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

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$${quality}\:{of}\:{questions}\:{you}\:{post}\:{really}\:{eye}\:{opener} \\ $$$${and}\:{activate}\:{sleeping}\:{cells}\:{of}\:{brain}\:{to}\:{get}\:{up} \\ $$$${and}\:{do}\:{work}... \\ $$

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