Eliminate θ\thetaθ from the equations
4x=cos(3θ)+3cosθ;4y=3sinθ−sin(3θ)4x=cos(3\theta)+3cos\theta; 4y=3sin\theta-sin(3\theta)4x=cos(3θ)+3cosθ;4y=3sinθ−sin(3θ)
Solution:
cos(3θ)=4cos3θ−3cosθcos(3\theta)=4cos^3\theta-3cos\thetacos(3θ)=4cos3θ−3cosθ
sin(3θ)=3inθ−4cos3θsin(3\theta)=3in\theta-4cos^3\thetasin(3θ)=3inθ−4cos3θ
4x=4cos3θ−3cosθ+3cosθ4x=4cos^3\theta-3cos\theta+3cos\theta4x=4cos3θ−3cosθ+3cosθ
4y=3sinθ−3sinθ+4sin3θ4y=3sin\theta-3sin\theta+4sin^3\theta4y=3sinθ−3sinθ+4sin3θ
x=cos3θx=cos^3\thetax=cos3θ
y=sin3θy=sin^3\thetay=sin3θ
x13=cosθx^{\frac{1}{3}}=cos\thetax31=cosθ
y13=sinθy^{\frac{1}{3}}=sin\thetay31=sinθ
cos2θ+sin2θ=1cos^2\theta+sin^2\theta=1cos2θ+sin2θ=1
Aswer:
x23+y23=1.x^{\frac{2}{3}}+y^{\frac{2}{3}}=1.x32+y32=1.
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