Triangle inequality for metric space where the metric is angles between vectors For $x,y \in S^1$, define $d(x,y)=cos^{-1}(\langle x,y \rangle)$ where $\langle x,y \rangle$ is the standard inner product.
Prove $d$ is a metric on $S^1$.
Because $-1 \leq \langle x, y \rangle \leq 1$ for $x,y \in S^1$, $0 \leq d(x,y) \leq \pi $ so positive definiteness is fine, and $d(x,y)=d(y,x)$ is also clear from the definition.
But I do not see how to show the triangle inequality, i.e.
$cos^{-1}(\langle x,y \rangle) \leq cos^{-1}(\langle x,z \rangle) + cos^{-1}(\langle z,y \rangle)$ for $x,y,z \in S^1$.
 A: The metric is defined as : for $x,y \in S^1$, the unit circle in the $\mathbb{R}^2 = \mathbb{C}$, as the smaller of the two angles that $x$ and $y$ make with the origin: if $x=e^{it}, y= e^{iu}$ ,with $t,u \in [0,2\pi)$, then $d(x,y) = \min(|t-u|, 2\pi - |t-u|)$. This is the same as your inner product formula, as $$\langle x,y\rangle = \langle (\cos(u) ,\sin(u)), (\cos(t) ,\sin(t)\rangle  =\cos u \cos t + \sin u \sin t = \cos(u-t)$$
The triangle inequality $d(x,z) \le d(x,y) + d(y,z)$ we only need to verify for three distinct points. There are some cases to consider. By rotating the circle we can move $x$ to $(1,0)$, keeping all distances. If $z$ lies in the first quadrant, then if $y$ lies on the short arc from $x$ to $z$, (draw pictures), then $d(x,y) + d(y,z) = d(x,z)$. The same happens if $z$ lies in the second (left upper) quadrant and $y$ lies ion the short arc again. If $y$ lies elsewhere, if $x=(1,0)$ lies on the short arc from $y$ to $z$, then $d(y,x) + d(x,z)  = d(y,z)$ and so $d(x,z) = d(y,z) - d(x,y) < d(x,y) + d(y,z)$ etc.
Similarly we can check all cases in the $\min$-formula (depending on which of the two terms is the smaller one), but this is tedious as well. 
