Sine Inequality/Proof 1

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 $x \ge 0$.

$\Box$

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

\(\ds \map {g'} x\)

\(=\)

\(\ds 2 x - 2 \sin x \cos x\)

Chain Rule for Derivatives and Derivative of Sine Function

\(\ds \)

\(=\)

\(\ds 2 x - \sin 2 x\)

Double Angle Formula for Sine

\(\ds \)

\(\ge\)

\(\ds 0\)

for $x \ge 0$, as shown above

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

$\blacksquare$