How may one show that $|\sin x-\sin y|\le |x-y|$?

Question

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?

Answer 1

Note

$$|\sin x- \sin y |=|2\cos\frac{x+y}2\sin\frac{x-y}2 |\le |2\sin\frac{x-y}2 |\le |x-y|$$

Answer 2

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.

Answer 3

$$ \begin{aligned} |\sin(x)-\sin(y)|&=\left|\int_y^x \cos(t)dt\right|\\&\leqslant\int_{\min(x,y)}^{\max(x,y)}\underbrace{|\cos(t)|}_{\leqslant 1}dt\\&\leqslant\max(x,y)-\min(x,y)\\&=|x-y| \end{aligned}$$

Answer 4

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 $\sin t$).