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.
Best Answer
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.