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How can we transform these parametric equations to Cartesian form?

$$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$

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What is the result of $x^2+y^2$? –  Sigur Dec 2 '12 at 0:38
    
$x^2+y^2 = r^2$. that was a hint? –  Rakisbro Dec 2 '12 at 0:41
    
Yes, it is constant 1, so the curve is contained on the unitary circle. Try to determine the end points. –  Sigur Dec 2 '12 at 0:42
    
I understand Sigur, thanks! –  Rakisbro Dec 2 '12 at 0:50

2 Answers 2

up vote 3 down vote accepted

If $-\pi\leq t\leq \pi$ then $-\pi/2\leq t/2\leq \pi/2$. Also $x^2+y^2=1$.

Here is the animated curve for $0\leq t\leq \pi$. Try to imagine what happens for $t$ negative.animated curve

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$$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$

$$x^2+y^2=(\sin \frac{t}{2})^2+(\cos \frac{t}{2})^2=1$$ so $$x^2+y^2=1$$ is equation of some circle

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