Since $\displaystyle\cot(2a) = \frac{\cos(2a)}{\sin(2a)}$, you would have (assuming you know the addition formulas for sines and cosines):
$$\begin{align*}
\cos(2a) &= \cos(a+a) = \cos(a)\cos(a) - \sin(a)\sin(a)\\
&= \cos^2(a) - \sin^2(a);\\
\sin(2a) &= \sin(a+a) = \sin(a)\cos(a) + \cos(a)\sin(a)\\
&= 2\sin(a)\cos(a),
\end{align*}$$
and therefore
$$\begin{align*}
\cot(2a) &= \frac{\cos(2a)}{\sin(2a)} = \frac{\cos^2(a) - \sin^2(a)}{2\sin(a)\cos(a)}\\
&= \frac{1}{2}\left(\frac{\cos^2(a)}{\sin(a)\cos(a)}\right) - \frac{1}{2}\left(\frac{\sin^2(a)}{\sin(a)\cos(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cos(a)}{\sin(a)} - \frac{\sin(a)}{\cos(a)}\right)\\
&= \frac{1}{2}\left(\cot(a) - \tan(a)\right)\\
&= \frac{1}{2}\left(\cot(a) - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a)}{\cot(a)} - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a) - 1}{\cot (a)}\right).
\end{align*}$$
P.S. Now, as it happens, I don't know the formulas for double angles, nor most identities involving tangents, cotangents, etc. I never bothered to memorize them. What I know are:
- The definitions of tangent, cotangent, secant, and cosecant in terms of sine and cosine;
- That sine is odd ($\sin(-x) = -\sin(x)$) and cosine is even ($\cos(-x)=\cos(x)$);
- The addition formulas for sine and cosine;
- The values of sine and cosine at $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, and $90^{\circ}$.
(I can derive $\sin^2\theta + \cos^2\theta = 1$ from the above, but in all honesty that one comes up so often that I do know it as well). I do not know the addition or double angle formulas for tangents nor cotangents, so the above derivation was done precisely "on the fly", as I was typing. I briefly thought that I might need to $\cos(2a)$ with one of the following equivalent formulas:
$$\cos^2(a)-\sin^2(a) = \cos^2(a) + \sin^2(a) - 2\sin^2(a) = 1 - 2\sin^2(a)$$
or
$$\cos^2(a) - \sin^2(a) = 2\cos^2(a) - \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1,$$
if the first attempt had not immediately led to a formula for $\cot(2a)$ that involved only $\cot(a)$ and $\tan(a) = \frac{1}{\cot(a)}$.
We give a brief sketch on how to introduce trigonometric functions through series. We will assume that basic properties of series have been proved. Yes, we need to define $\pi$. We will define it in a way close in spirit to what you did. Our aim is to set up some useful machinery, enough to fully answer your question about $\sin(\pi/2)$.
The Pythagorean Identity: By properties of series, the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. Now differentiate $\sin^2 x+\cos^2 x$. We get $2\sin x\cos x-2\cos x\sin x$. This is $0$, so $\sin^2 x+\cos^2 x$ is constant. It is easy to show that the constant is $1$.
The Addition Laws: We prove the Addition Laws for sine and cosine. Fix $y$, and let
$$\begin{align}f(x)&=\sin(x+y)-\left(\sin x\cos y+\cos x\sin y\right) \\
g(x)&=\cos(x+y)-\left(\cos x\cos y-\sin x\sin y\right)\end{align}$$
It is easy to verify that $f'(x)=g(x)$ and $g'(x)=-f(x)$. Now let $S(x)=f^2(x)+g^2(x)$. We find, as in an earlier calculation, that $S'(x)=0$. So $S(x)$ is constant. But easily by substitution $S(0)=0$. It follows that $S(x)$ is identically $0$, and the Addition Laws for sine and cosine follow.
On Defining $\pi$: We can use the series to estimate $\cos 1$, and show that it is strictly between $1/2$ and $13/24$. From this it follows by $\cos 2x=2\cos^2 x-1$ that $\cos 2$ is negative. By continuity, there is a smallest positive number $q$ such that $\cos q=0$. Define $\pi$ to be $2q$.
The Value of $\sin(\pi/2)$: Since $\cos q=0$, we have by the definition of $\pi$ that $\cos(\pi/2)=0$. By the Pythagorean Identity, we have $\sin(\pi/2)=\pm 1$. But it cannot be $-1$, for since $\pi/2$ is the smallest positive zero of $\cos x$, the sine function is increasing from $0$ to $\pi/2$.
Remark: It looks like a lot of work, but once the basics have been established, the rest goes smoothly. We proved the full Addition Laws instead of the fragment needed for $\sin(\pi/2)$ to make it clear that other standard facts are not difficult to derive.
Best Answer
Here is a geometric interpretation that is easy to remember: the unit circle is parametrized by $(\cos t, \sin t)$ and hence its tangent vector is orthogonal to the position vector. Rotating the position vector by 90 degrees gives you $(-\sin t, \cos t)$ and so $\cos'=-\sin$ and $\sin'=\cos$.
This argument has a simple interpretation in terms of complex numbers. $\exp(it) = \cos t + i\,\sin t$ and so $i \exp(it) = \exp(it)'= \cos' t + i\,\sin' t$. But $i \exp(it) = -\sin t +i\,\cos t$.
These two arguments are the same, really.