I think I saw the inequality $$|\sin x-\sin y|\le |x-y|$$ somewhere recently.
I've been trying to see why it could be true but so far have been unable to come up with any success.
Could someone show a way to prove this?
Thanks.
PS. Intuitively, this seems to make sense since an arc is always bigger than its sine in magnitude. But how to show this rigorously?
The way to make this rigorous is via the Mean Value theorem. For any $x<y\in \mathbb{R}$ there is a $c\in (x,y)$ s.t. $$\sin(x)-\sin(y) = \cos(c)(x-y).$$ But since $|\cos(z)|\leq 1$ for any $z\in \mathbb{R}$, you can take the absolute value and immediately get the result.
$$ \begin{aligned} |\sin(x)-\sin(y)|&=\left|\int_y^x \cos(t)dt\right|\\&\leqslant\int_{\min(x,y)}^{\max(x,y)}\underbrace{|\cos(t)|}_{\leqslant 1}dt\\&\leqslant\max(x,y)-\min(x,y)\\&=|x-y| \end{aligned}$$
Note
$$|\sin x- \sin y |=|2\cos\frac{x+y}2\sin\frac{x-y}2 |\le |2\sin\frac{x-y}2 |\le |x-y|$$
Using the identity $$\sin x - \sin y = 2 \sin \frac{x-y}2 \cos \frac{x+y}2$$ and the fact that $\sin$ is an odd function it will indeed follow from inequality $$\sin t \le t \mathrm{~~for~~} t \ge 0$$
There are many analytical ways of solving this, my favourite being the comparison of derivatives.
Geometrically you can argue that $\sin t$ (blue segment) is shorter than red which is shorter than an arc of length $t$:
(This works for $t < \pi/2$, what includes $t=1$ which is the maximal value of $\sin t$).
