Proof of the derivative of sine using vectors

Question

I have found a nice proof that the derivative of sine is cosine that uses vectors, but I'm struggling to properly understand it. As $\theta$ increases, the point $(\cos \theta, \sin \theta)$ moves around the unit circle. At any given point, the direction of travel is perpendicular to the point's position, and so the rate of change is proportional to $(-\sin \theta, \cos \theta)$. If $\theta$ is measured in radians, then the proportionality constant is equal to $1$, and we get that $$ \frac{d}{d\theta}(\cos\theta,\sin\theta)=(-\sin \theta, \cos \theta) $$ I'm finding this proof difficult to understand because I am unfamiliar with vector calculus. What does the length of the tangent vector represent? E.g. if $\theta$ was measured in degrees, then the rate of change would be $$ \frac{\pi}{180}(-\sin\theta,\cos\theta) $$ So presumably the length of the tangent vector would change, but I don't exactly grasp how. For derivatives, $\frac{dy}{dx}$ tells us the gradient of the tangent to a particular point, but the actual length of the tangent is infinite. So I suppose I am asking about the difference between a tangent and a tangent vector.