Parametrization of circle So here is a parametric equation for circle where
$$x=\cos t,\quad y=\sin t$$
but this only traces out a circle when $t$ goes from $0$ to $2\pi$.
If I were to trace a circle with the restriction of $0$ to $\pi$, it would be ($\cos 2t$, $\sin 2t$).
However, why is it not $(\cos \frac{t}{2},\sin \frac{t}{2})$ since essentially here we are just dividing the original $0$ to $2\pi$ bounds by $2$, giving us $\frac{t}{2}$ is between $0$ and $\pi$? (I directly substituted $t$ with $\frac{t}{2}$ into the $\cos t$ and $\sin t$, but why would that not work?)
 A: As you've pointed out, when you use the parameterization $(\cos(t),\sin(t))$ you need $0\le t \le 2\pi$.  In other words, whatever you input into cosine and sine needs to vary between $0$ and $2\pi$ in order to trace a full circle.  If you change the input to $t/2$ you still need that input to vary between $0$ and $2\pi$.  In other words, you need $$0\le \frac{t}{2}\le 2\pi \implies 0 \le t \le 4\pi.$$  With $2t$ as the input instead you still need that input to vary between $0$ and $2\pi$, which gives you $$0 \le 2t \le 2\pi \implies 0 \le t \le \pi.$$  With whatever form you chose for your input, that input has to vary between $0$ and $2\pi$ to get a whole circle the way you've been doing it, but the domain for $t$ will change depending on the choice of input.
Edit to address comment:
You have to be a little bit careful here.  If you have the parameterization $(\cos(t),\sin(t))$, then you are correct in saying that $0 \le t \le 2\pi$ is the same as saying $0 \le t/2 \le \pi$.  However, that's all for the same $t$ and the same parameterization.  So, it's fine to say that $(\cos(t),\sin(t))$ traces out a full circle as $t/2$ varies from $0$ to $\pi$ because that's the same as saying that $t$ varies from $0$ to $2\pi$, but if you want to replace $t$ with $t/2$ in the parameterization you need to actually make that change everywhere.  So,
\begin{align}(\cos(t),\sin(t))&\longrightarrow (\cos(\color{red}{t/2}),\sin(\color{red}{t/2}))\\
0\le t \le 2\pi &\longrightarrow 0 \le \color{red}{\frac{t}{2}} \le 2\pi \implies 0 \le t\le 4\pi\\
\text{or equivalently:}\quad 0 \le \frac{t}{2} \le \pi &\longrightarrow 0 \le \frac{\color{red}{t/2}}{2}\le \pi \implies 0 \le t \le 4\pi
\end{align}
