How do you prove that a particle with the position $(\cos t, \sin t)$ has unit speed?

Question

Consider a particle that's motion is given by the vector function $$ S(t)=(\cos t,\sin t) \, , $$ where $t$ is measured in radians. How do you prove that the particle has unit speed, i.e. $|S'(t)|=1$? At first, I thought this would be easy, since $\cos t$ and $\sin t$ are simply the $x$- and $y$-coordinates of a particle that has traced an arc of $t$ units anticlockwise around the unit circle. This means that between two points $S(t+h)$ and $S(t)$, the particle must have travelled a distance of $h$ units. Hence, $$ |S'(t)|=\lim_{h \to 0}\frac{h}{h}=1 \, . $$ However, I'm unsure if this solution is actually correct. If I understand correctly, $|S'(t)|$ is the limit of the displacement from the initial position, i.e. $$ |S'(t)| := \lim_{h \to 0}\frac{\sqrt{(\cos(t+h)-\cos t)^2+(\sin(t+h)-\sin t)^2}}{h} \, . $$ In other words, although the particle will have travelled a distance of $h$ units in going from $S(t)$ from $S(t+h)$, its displacement would be something else. However, according to Wolfram Alpha, the above limit does not even exist. So it seems that I have some misunderstanding of what $|S'(t)|$ actually means, but I'm unsure what it is.

Answer 1

Your mistake is in dropping the absolute value.

$$ |S'(t)| := \lim_{h \to 0}\frac{\sqrt{(\cos(t+h)-\cos t)^2+(\sin(t+h)-\sin t)^2}}{|h|} =1. $$

Answer 2

$$\vec{r}(t)=\hat{i}\cos(t)+\hat{j}\sin(t)$$ $$\vec{v}(t)=\frac{d\vec{r}}{dt}=-\hat{i}\sin(t)+\hat{j}\cos(t)$$ $$|\vec{v}|=\sqrt{[-\sin(t)]^2+\cos^2(t)}=1(\therefore)$$