Modulus of Sine of x Less Than or Equal To Absolute Value of x

Theorem

Let $x$ be a real number.

Then:

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

Proof

Clearly the inequality holds if $x = 0$.

Take $x \ne 0$.

From the Mean Value Theorem and Derivative of Sine Function, there exists $c \in \R$ such that:

$\ds \frac {\sin x - \sin 0} {x - 0} = \cos c$

so:

$\sin x = x \cos c$

Then we have:

\(\ds \size {\sin x}\)

\(=\)

\(\ds \size {x \cos c}\)

\(\ds \)

\(\le\)

\(\ds \size x\)

as required.

$\blacksquare$