Cosine Inequality

Theorem
$1 - \dfrac {x^2} 2 \le \cos x$

for all $x \in \R$.

Proof
Let $f \left({x}\right) = \cos x - \left({1 - \dfrac {x^2} 2 }\right)$.

By Derivative of Cosine Function, $f' \left({x}\right) = x - \sin x$.

From Sine Inequality, we know $\sin x \le x$ for $x \ge 0$.

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

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

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

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

By Cosine Function is Even, $f \left({x}\right)$ is an even function.

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