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I am supposed to find a Cartesian equation by eliminating the parameters of

$x = \sin \left(\dfrac{1}{2}\theta \right), \qquad y = \cos \left(\dfrac{1}{2}\theta \right)$

I won't even go through my work since it is so far off but I am suppose to get $x^2 + y^2 = 1$ but I get something with $\arccos$ and such which doesn't make any sense.

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The equation $x^2+y^2=1$ works. How can we tell? Experience: if you have a sine and a cosine of the same quantity (in this case, of $\frac{1}{2}\theta$), then you can always "get rid" of the dependence on the quantity by squaring and adding; no matter what the quantity is, you always get $\sin^2(\text{quantity}) + \cos^2(\text{quantity}) = 1$.

But let's say you don't want to try things to see how the parameter can be eliminated. In this case, you can simply solve for the parameter in each equation: $\begin{align*} x&=\sin\left(\frac{1}{2}\theta\right)\\ \arcsin(x) &= \frac{1}{2}\theta\\ 2\arcsin(x) &=\theta;\\ y &= \cos\left(\frac{1}{2}\theta\right)\\ \arccos(y) &= \frac{1}{2}\theta\\ 2\arccos(y) &= \theta. \end{align*}$ Therefore, $x$ and $y$ will satisfy $2\arcsin(x) = 2\arccos(y)$ or equivalently, $\arcsin(x) = \arccos(y).$

The problem is that this equation is ugly; arcsine and arccosine are annoying functions. Better to try to get rid of them. One way to do that is to first take sines on both sides, to get $x = \sin(\arccos(y)).$ Now, what is $\sin(\arccos(y))$? Well, since $\sin^2z + \cos^2z = 1$ (there is again!), then $\sin^2 z = 1-\cos^2 z$, so $|\sin z| = \sqrt{1-\cos^2(z)}$. Letting $z=\arccos(y)$, we get $\sin(\arccos(y)) = \pm\sqrt{1 - \cos^2(\arccos(y))} = \pm\sqrt{1-y^2}.$ So our equation becomes $x = \pm\sqrt{1-y^2}.$ Again, the $\pm$ is annoying. We can get rid of it by squaring both sides, so we get $x^2 = 1-y^2$ which can then be rewritten as $x^2 + y^2 = 1.$ Now, this is a nice equation; we know exactly what it is (circle of radius $1$ centered at the origin), and it does not involve any annoying functions like inverse trigonometric functions. So this is a good stopping point.

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    @ArturoMagidin True. I guess it can get wierd.2012-06-22
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Remember that for all $z$, we have $\cos^2(z) + \sin^2(z) = 1$

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    But I don't know why they are being added.2012-06-21
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This is JUST to clear what Marvis noted.

$x=\sin(\frac{1}{2}\theta)$ $\longrightarrow$ $x^2=\sin^2(\frac{1}{2}\theta)$ ,

$y=\cos(\frac{1}{2}\theta)$ $\longrightarrow$ $y^2=\cos^2(\frac{1}{2}\theta)$

Checking an elementary Mathematics book you'll find there is an trigonometric equation (See Marvis's answer). Now if you get $\frac{1}{2}\theta=z$ or $x$ or $y$ or every name you want and by adding above conclusions $x^2+y^2=1$. Isn't this enough? :-)

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    *Eliminating the parameter* means **finding** some equation involving $x$ and $y$ that expresses a relationship between them that does not depend on the parameter. In this case, one such equation is $x^2+y^2=1$. They get added because *that* eliminates the parameter. Alternatively, solve for the parameter in terms of $x$ and $y$, $\theta = 2\arcsin(x)$, $\theta=2\arccos(y)$, and then set them equal: $2\arcsin(x)=2\arccos(y)$; hence $\arcsin(x)=\arccos(y)$ is one such equation. This equation can be converted into $x^2+y^2=1$ by suitable manipulations.2012-06-21
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Just take,

$ x^2 = \sin^2\left(\frac{\theta}{2}\right) $ and $y^2 = \cos^2\left(\frac{\theta}{2}\right) ,$

$ cos^2(\phi) + sin^2(\phi) = 1$

$\implies x^2 + y^2 = 1$ That's it! $\LaTeX$