Sine inequality: How to prove that $|\sin(x)| \le |x|$ for $ x \in \mathbb{R}$ [closed]

calculus trigonometry inequality

If $x\in\mathbb R$, $\bigl|\sin'(x)\bigr|=\bigl|\cos(x)\bigr|\leqslant1$. Therefore, by the mean value theorem,$$\bigl|\sin(x)\bigr|=\bigl|\sin(x)-\sin(0)\bigr|\leqslant1\times|x-0|=|x|.$$


Note that the sine of theta is less than the arc-length. Pretty standard approach to evaluating the limit:

$$\lim_{x\to0}\frac{\sin(x)}x$$

enter image description here

This proof is only valid for $0<x<\pi/2$. For angles beyond this, simply note that

$$|\sin(x)|\le1$$

and

$$\sin(x)\le x\implies|\sin(-x)|\le|-x|$$

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