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Question Number 75302 by vishalbhardwaj last updated on 09/Dec/19
If θ is eleminated from the equation x = a cos(θ−α) and y = b cos(θ−β) then prove that (x^2 /a^2 ) + (y^2 /b^2 ) − ((2xy)/(ab)) cos(α−β) = sin^2 (α−β) ?
$$\mathrm{If}\:\theta\:\mathrm{is}\:\mathrm{eleminated}\:\mathrm{from}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}\:=\: \\ $$$${a}\:{cos}\left(\theta−\alpha\right)\:\mathrm{and}\:{y}\:=\:{b}\:{cos}\left(\theta−\beta\right)\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}{xy}}{{ab}}\:{cos}\left(\alpha−\beta\right)\:=\:{sin}^{\mathrm{2}} \left(\alpha−\beta\right)\:? \\ $$
Commented by vishalbhardwaj last updated on 10/Dec/19
please solve this
$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this} \\ $$
Answered by peter frank last updated on 03/Jan/20
x=acos (θ−α) y=bcos (θ−β) θ=cos^(−1) ((x/a))+α θ=cos^(−1) ((y/b))+β α−β=cos^(−1) ((x/a))−cos^(−1) ((y/b)) cos^(−1) ((x/a))=P Q=cos^(−1) ((y/b)) sin P=(√(1−(y^2 /b^2 ))) cos P=(√(1−(x^2 /a^2 ))) α−β=P−Q cos (α−β)=cos (P−Q) =xy+(√((1−(x^2 /a))(1−(y^2 /b^2 )))) α−β=P−Q sin (α−β)=sin (P−Q)w sin^2 (α−β)=((2xy)/(ab))(√((1−(x^2 /a^2 ))(1−(y^2 /b^2 )))) sin^2 (α−β)=((2xy)/(ab))cos (α−β) +(x^2 /a^2 )+(y^2 /b^2 )
$${x}={a}\mathrm{cos}\:\left(\theta−\alpha\right)\:\: \\ $$$${y}={b}\mathrm{cos}\:\left(\theta−\beta\right) \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)+\alpha \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{y}}{{b}}\right)+\beta \\ $$$$\alpha−\beta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)−\mathrm{cos}^{−\mathrm{1}} \left(\frac{{y}}{{b}}\right) \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)={P}\:\:\:\:\:{Q}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{y}}{{b}}\right) \\ $$$$\mathrm{sin}\:{P}=\sqrt{\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:{P}=\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$$\alpha−\beta={P}−{Q} \\ $$$$\mathrm{cos}\:\left(\alpha−\beta\right)=\mathrm{cos}\:\left({P}−{Q}\right) \\ $$$$={xy}+\sqrt{\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}}\right)\left(\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)} \\ $$$$ \\ $$$$\alpha−\beta={P}−{Q} \\ $$$$\mathrm{sin}\:\left(\alpha−\beta\right)=\mathrm{sin}\:\left({P}−{Q}\right){w} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\alpha−\beta\right)=\frac{\mathrm{2}{xy}}{{ab}}\sqrt{\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\alpha−\beta\right)=\frac{\mathrm{2}{xy}}{{ab}}\mathrm{cos}\:\left(\alpha−\beta\right)\:\:\:+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$ \\ $$

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