x=sin⁡(θ2),y=cos⁡(θ2),−π≤θ≤π (a) Eliminate the parameter to find a Cartesian equation of the curve. and how does thecurve look

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Liyana Mansell · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Consider the following parametric equations,  x=sin⁡(θ2) and y=cos⁡(θ2), −π≤θ≤π  To eliminate θ, square x and y both sides and add them  x2=sin2⁡(θ2)  y2=cos2⁡(θ2)  x2+y2=sin2⁡(θ2)+cos2⁡(θ2)  x2+y2=1  The equation x2+y2=1 represents a circle of radius. Given that the perimeter θ varies from −π to π  Substitute θ=−π in the perimetric equations  x=sin⁡(θ2) and y=cos⁡(θ2)  x=sin⁡(−π2)=−1  y=cos⁡(−π2)=0  So the point corresponding to θ=−π is (−1,0)  Substitute θ=π in the perimetric equations  x=sin⁡(θ2) and y=cos⁡(θ2)  x=sin⁡(−π2)=−1  y=cos⁡(−π2)=0  So the point corresponding to θ=−π is (−1,0)  Substitute θ=π in the perimetric equations  x=sin⁡(θ2) and y=cos⁡(θ2)  x=sin⁡(π2)=1  y=cos⁡(π2)=0  So, the point corresponding to θ=π is (1,0)  Hence, the curve is a senincircle with origin (0,0) radius 1.  As θ increases from −π to π, the point traces semicircles in clockwise direction, storting from the point (−1,0) to the point (1,0)

x =\sin \left(\frac{\theta}{2}\right), y = \cos \left(\frac{\theta}{2}\right), -\pi \leq \theta \leq \pi (a) Eliminate the parameter to find a Cartesian equation of the curve. and how does thecurve look

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