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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 $x^2+y^2$?2012-12-02
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    $x^2+y^2 = r^2$. that was a hint?2012-12-02
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    Yes, it is constant 1, so the curve is contained on the unitary circle. Try to determine the end points.2012-12-02
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    I understand Sigur, thanks!2012-12-02

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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