Parametrization of a circle, why is it $(\sin t, \cos t)$ instead of $(\sin^2t,\cos^2t)$?

Question

The equation of a circle is $x^2 + y^2 = r^2$.

Then why is the parameterization of a circle $(r\cos(t), r\sin(t))$ and not $(r\cos^2(t), r\sin^2(t))$?

I thought what I'm supposed to do here is to eliminate $x$ and $y$ and write them in a way where I can represent them both with $t$ and then put it back into the equation? If I do that, I get

$$x = r\cos(t)$$ $$y = r\sin(t)$$

then if I put it back into the equation of the circle I get them both squared. Shouldn't the result then be $(\sin^2t,\cos^2t)$?

What's wrong here?

Answer 1

If $x = r \cos t$, $y = r \sin t$, then $$x^2+y^2 = r^2 \cos^2 t + r^2 \sin^2 t = r^2 (\cos^2 t + \sin^2 t)$$ which is indeed $r^2$.

If $x = r \cos^2 t$, $y = r \cos^2 t$, then $$x^2 + y^2=(r \cos^2 t)^2 + (r \sin^2 t)^2 = r^2 \cos^4 t + r^2 \sin^4 t$$ This is not equal to $r^2$ in general.

Answer 2

If $(x, y) = (r\cos^2(t), r\sin^2(t)) $ then $x+y=r$ which is the equation of a straight line, not a circle.

Answer 3

When we use the notation $(x,y)$ with regards to Cartesian co-ordnates we mean that the $x$-coordinate is $x$ and the $y$-coordiante is $y$ (unsurprisingly). If we have $x$ and $y$ expressed parametrically then we can replace each variable by the equation in the given parameter it is equivalent to; eg in your case we'd replace $x$ by $r\cos{t}$.