Sine Inequality

Theorem

 * $\size {\sin x} \le \size x$

for all $x \in \R$.

Proof
Let $f \paren x = x - \sin x$.

By Derivative of Sine Function:
 * $f' \paren x = 1 - \cos x$

From Real Cosine Function is Bounded we know $\cos x \le 1$ for all $x$.

Hence $f \paren x \ge 0$ for all $x$.

From Derivative of Monotone Function, $f \paren x$ is increasing.

By Sine of Zero is Zero, $f \paren x = 0$.

It follows that $f \paren x \ge 0$ for all $x \ge 0$.

Now let $g \paren x = x^2 - \sin^2 x$.

From Derivative of Monotone Function, $g \paren x$ is increasing for $x \ge 0$.

By Sine of Zero is Zero, $g \paren x = 0$.

It follows that $g \paren x \ge 0$ for all $x \ge 0$.

Observe that $g \paren x$ is an even function.

This implies $g \paren x \ge 0$ for all $x \in \R$.

Finally note that $\sin^2 x \le x^2 \iff \size {\sin x} \le \size x$.