In a math book I am reading, the author mentions that the first quadrant of a circle can be defined by the parametric equations \begin{align*} x(u)&=\cos{u}\\ y(u)&=\sin{u}, \end{align*} for $0\leq u\leq\frac{\pi}{2}$. The author then goes on to say that by setting $t=\tan(u/2)$, one can derive the alternate representation \begin{align*} x(t)&=\frac{1-t^2}{1+t^2}\\ y(t)&=\frac{2t}{1+t^2}, \end{align*} for $0\leq t\leq 1$.
Let's consider the function $x$ for a moment. I understand, that if $t=\tan(u/2)$, then $u=2\arctan{t}$, and $\cos{u}=\cos(2\arctan{t})=(1-t^2)/(1+t^2).$ What I can't see is how this is equal to $x(t)$. It seems like it should be equal to $x(2\arctan{t})$! When I actually try to compute $x(t)=x(\tan(u/2))$, I inevitable get stuck and cannot arrive at $(1-t^2)/(1+t^2)$.
Can someone please help me to connect the dots here?
I think your confusion is coming from the fact that $x\mapsto x(u)$ and $x\mapsto x(t)$ are not the same function, despite having the same name. We can be a bit more precise:
Consider $x(u) = \cos u$, and let $u = 2\arctan t$. Then $$x(u) = x(2\arctan t) = \cos(2\arctan t) = \frac{1-t^2}{1+t^2}.$$ Now, let's define $x_1(t)$ by $$x_1(t) = x(2\arctan t) = \frac{1-t^2}{1+t^2}.$$ In the same way, we can start with $y(u)$ and get to $$y_1(t) = y(2\arctan t) = \frac{2t}{1+t^2}.$$ So the functions $x,y$ and $x_1,y_1$ are in fact distinct functions, although we can think of $x_1$ and $y_1$ as compositions of $x$ and $y$ with $2\arctan t$.
The important thing here is that $$x_{1}^2(t)+y_{1}^2(t) = \left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 = \frac{1 -2t^2 + t^4 + 4t^2}{1+2t^2 + t^4} = \frac{1+2t^2+t^4}{1+2t^2+t^4} = 1,$$ and so this parameterization does in fact give the unit circle.