Why is it that $x=\cos(t)$ and $y=\sin(t)$?

Question

I'm supposed to find the curve of intersection of $z=x^2-y^2$ and $x^2+y^2=1$. I usually go about this by parametrization. I set $x=t$ and plug and chug the other variables. However that won't work easily in this situation.

My instructor recommended I use $x=\cos(t)$ and $y=\sin(t)$ instead. I'm curious as to how and why this will work, since I know that my method of parametrization doesn't change the values of the original variables, but this one seems to.

So why does using $x=\cos(t)$ and $y=\sin(t)$ work?

Answer 1

If you set x = cos(t) and y = sin(t), then the equality x$^2$ + y$^2$ = 1 is always fulfilled, due to the Pythagorean identity sin$^2$(x) + cos$^2$(x) = 1. Therefore, since the second equation is always true given the circumstances, you can simply plug in the values of x and y into the first equation and that will give you your desired curve, which in this case is z = x$^2$ - y$^2$, or z = cos$^2$(t) - sin$^2$(t) or, according to the double angle formula: z = cos(2t)

Answer 2

There is a trig identity you may recall, $cos^{2}\theta + \sin^{2}\theta=1$. This answers your direct question since you can match up $x^2 + y^2 =1$

Answer 3

This is literally the definition of $\sin$ and $\cos$. The $\sin$ of a number $t$ is the $y$-coordinate of a point $t$ radians along the unit circle, measured in a counterclockwise direction from $(1,0)$. The $\cos$ of $t$ is that same point's $x$-coordinate. Thus, as $t$ increases from $0$, the point $(\cos t,\sin t)$ simply tracks motion around the circle, by definition.

You may have seen $\sin$ and $\cos$ defined some other way, perhaps in the context of right triangles? Showing that the two definitions are equivalent is a good exercise. These functions can also be defined other ways, such as in terms of certain differential equations. The unit circle definition is a good one, though, for many purposes.