Finding Cartesian equations from parametric equations 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.
 A: 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.
A: Remember that for all $z$, we have $$\cos^2(z) + \sin^2(z) = 1$$
A: 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? :-)
A: 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!
$$\LaTeX$$
