Is this a valid proof of the derivatives of the trigonometric functions? For the sake of this proof, the trigonometric functions $\cos$ and $\sin$ are defined as the coordinates of a point on the unit circle, rather than any of the modern analytic definitions.
Let $\vec x(\theta)$ be a function giving the position of a point on the unit circle as a function of the angle $\theta$, that is, $\vec x(\theta) = \left(
\begin{array}{c}
\cos(\theta)\\
\sin(\theta)\\
\end{array}
\right)$. Decompose $\vec x'(\theta)$ into a colinear and an orthogonal component with $\vec x(\theta)$. If the colinear component is non-null, then on the assumption that $\vec x'(\theta)$ is continuous (could this be demonstrated?), it would be non-null over some small interval (right?), and then by taking the integral over that interval, we would have that $\vec x(\theta)$ either moves away from or towards the center of the origin, which is contrary to our hypotheses. Thus the colinear component is always null, and $\vec x'(\theta)$ is always tangent to the circle.
But for any vector $\left(
\begin{array}{c}
a\\
b\\
\end{array}
\right)$, a vector is orthogonal to it if and only if it is colinear with $\left(
\begin{array}{c}
-b\\
a\\
\end{array}
\right)$. Therefore $\vec x'(\theta) = k\left(
\begin{array}{c}
-\sin(\theta)\\
\cos(\theta)\\
\end{array}
\right)$ and we have $\cos' = -k\sin$, $\sin' = k\cos$, for some $k$.
There are gaps in this proof (in italics), but are there ways of filling them in while remaining within the geometric definitions of the trigonometric functions? It would be pleasing to be able to prove the derivatives of the trig functions purely on the basis of the fairly obvious geometrical nature of circular motion.

A hint: In order to determine $k$ use the fact that $|\vec x(\theta)|^2\equiv1$.
