Cosine Inequality

Theorem

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

for all $x \in \R$.

Proof

Let $\map f x = \cos x - \paren {1 - \dfrac {x^2} 2}$.

By Derivative of Cosine Function:

$\map {f'} x = x - \sin x$

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

Hence $\map {f'} x \ge 0$ for $x \ge 0$.

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

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

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

By Cosine Function is Even, $\map f x$ is an even function.

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

$\blacksquare$