Answer:
We know that equation of a circle with origin as it's center is given by
[tex]x^{2}+y^{2}=r^{2}\\\\\therefore x^{2}+y^{2}=2^{2}\\\\(\frac{x}{2})^{2}+(\frac{y}{2})^{2}=1\\\\[/tex]
Comparing with [tex]sin^{2}(\theta )+cos^{2}(\theta )=1[/tex] we get
[tex]\frac{x}{2}=sin(\theta )\\\\\therefore x=2sin(\theta )\\\\\frac{y}{2}=cos(\theta )\\\\\therefore y=cos(\theta )[/tex]
Since [tex]sin(\theta ),cos(\theta )[/tex] have a period of 2π but in the given question we need to increase the period to 4π thus we reduce the argument by 2
[tex]\therefore x=2sin(\frac{\theta }{2})\\\\y=2cos(\frac{\theta }{2})[/tex]