Sine Inequality

Theorem

 * $\left|{\sin x}\right| \le \left|{x}\right|$

for all $x \in \R$.

Proof
Let $f \left({x}\right) = x - \sin x$.

By Derivative of Sine Function, $f' \left({x}\right) = 1 - \cos x$.

From Boundedness of Sine and Cosine we know $\cos x \le 1$ for all $x$.

Hence $f' \left({x}\right) \ge 0$ for all $x$.

From Derivative of Monotone Function, $f \left({x}\right)$ is increasing.

By Sine of Zero is Zero, $f \left({x}\right) = 0$.

It follows that $f \left({x}\right) \ge 0$ for all $x \ge 0$.

Now let $g \left({x}\right) = x^2 - \sin^2 x$.

From Derivative of Monotone Function, $g \left({x}\right)$ is increasing for $x \ge 0$.

By Sine of Zero is Zero, $g \left({x}\right) = 0$.

It follows that $g \left({x}\right) \ge 0$ for all $x \ge 0$.

Observe that $g \left({x}\right)$ is an even function.

This implies $g \left({x}\right) \ge 0$ for all $x \in \R$.

Finally note that $\sin^2 x \le x^2 \iff \left|{\sin x}\right| \le \left|{x}\right|$.