The following problem is taken from 'Real Mathematical Analysis' by Pugh $2$nd edition, page $125,$ exercise $6.$
For $p,q \in [0,\pi/2)$ let $$d_s(p,q)= \sin |p-q|.$$ Use your calculus talent to decide whether $d_s$ is a metric.
My attempt:
Clearly for any $p,q \in [0,\pi/2),$ $d_s(p,q) \geq 0.$
If $p=q,$ then we have $d_s(p,q) = \sin |p - p| = 0.$
If $d_s(p,q) = 0,$ then $\sin |p-q| = 0.$ As sine function is injective on $[0,\pi/2),$ we have $|p - q| = 0,$ which implies that $p = q.$
For any $p,q \in \in [0,\pi/2),$ we have $d_s(p,q) = \sin |p - q| = \sin |q - p| = d_s(q,p).$
Let $p,q,r \in [0,\pi/2).$
Since $d_s(p,q) = d_s(q,p),$ without loss of generality, we assume that $p \geq q \geq r.$
Observe that
$$d_s(p,r) \leq d_s(p,q) + d_s(q,r)$$
$$\Leftrightarrow \sin(p-r) \leq \sin(p-q) + \sin(q-r)$$
$$\Leftrightarrow 2 \sin \left( \frac{p-r}{2} \right) \cos \left( \frac{p-r}{2} \right) \leq 2 \sin \left( \frac{p-r}{2} \right) \cos \left( \frac{p-2q +r}{2} \right)$$
$$\Leftrightarrow \cos \left( \frac{p-r}{2} \right) \leq \cos \left( \frac{p- 2q + r}{2} \right)$$
Since cosine function is decreasing and $\frac{p-r}{2}, \frac{p-2q+r}{2} \in [0,\pi/4),$ we have
$$\Leftrightarrow \frac{p-r}{2} \geq \frac{p-2q+r}{2}$$
$$\Leftrightarrow q \geq r.$$
Since the last inequality holds, we have $d_s(p,r) \leq d_s(p,q) + d_s(q,r).$
Hence, $d_s$ is a metric.
I am not very sure about triangle inequality part. Can anyone check my proof? Thanks.
It is easie. For $x\in [-\pi/2; \pi/2]$ we have $\sin |x|=|\sin x|$, therefore $\sin|p-r|=|\sin(p-r)|=|\sin(p-q)\cos(q-r)+\sin(q-r)\cos(p-q)|\leq |\sin(p-q)|+|\sin(q-r)|=\sin|p-q|+\sin|q-r|.$