Sine Inequality

Theorem

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

for all $x \in \R$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof 2
For $x=0$, the claim is clear.

If $\size x \ge 1$, the claim holds as:

Now, suppose $\size x < 1$

Then:

The last expression is $\le 1$, since: