Finding Cartesian equations from parametric

I am suppose to find a cartesian equation by eliminating the parameters of

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

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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@BabakSorouh: Okay, that's it. Please stop commenting on most of my answers with comments that have nothing but expressions of praise. Not only do some of them seem rather exaggerated, but, frankly, I'm finding it creepy. I'm feeling stalked, and I don't like it. – Arturo Magidin Jun 21 '12 at 22:02
I guess you're right, but he seems to write similar comments in most answers. – Pedro Tamaroff Jun 22 '12 at 13:03
@Peter: I only get notified when he posts them in mine, though... – Arturo Magidin Jun 22 '12 at 14:41
@ArturoMagidin True. I guess it can get wierd. – Pedro Tamaroff Jun 22 '12 at 15:32

Remember that for all $z$, we have $$\cos^2(z) + \sin^2(z) = 1$$

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I don't know what that weird a is. – user138246 Jun 21 '12 at 20:26
I give up ${}{}$ – user17762 Jun 21 '12 at 20:31
Because Jordan :) – user17762 Jun 21 '12 at 20:40
@Jordan: Dear Jordan, Marvis was trying to tell you that the equation $\cos^2(z)+\sin^2(z)=1$ can make your question solved. In fact, if you change your variable from $\frac{1}{2}\theta$ to $z$ then by using a very basic formula, you will get your answer as you want. This basic trigonometric formula is what Marviz noted you. – Babak S. Jun 21 '12 at 20:44
$x$ and $y$ are related because $x$ is the sine of a certain thing, and $y$ is the cosine of the same thing, and sine and cosine are closely related to one another, as you might guess from the names. – MJD Jun 21 '12 at 21:07

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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No you miraculously managed to skip the only part I do not understand. I already know every single thing stated in this thread except the one thing everyone is unwilling to tell me. Why is the y and x being added and why it is equal to one? – user138246 Jun 21 '12 at 21:03
@Jordan: $x$ and $y$ is not being added here!. The squares of them separately are being added. Don’t you want to reach $x^2+y^2=1$??? Don't we see an adding in the previous equation? – Babak S. Jun 21 '12 at 21:09
Yes I see adding, and that is why I am confused. Why are they being added together? – user138246 Jun 21 '12 at 21:12
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. – Arturo Magidin Jun 21 '12 at 21:25

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! $$\LaTeX$$

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