Use of inequality $1 - \cos (x) \leq x^2 /2$

Solution 1:

The bound $-1\le \cos(x)\le 1$ is correct, but rather a crude one.

We can easily obtain tighter bounds using the inequality from elementary geometry

$$|\sin(\theta)|\le |\theta| \tag 1$$

for all $\theta$, and the half-angle formula for the sine function

$$1-\cos(\theta)=2\sin^2(\theta/2) \tag 2$$

Squaring both sides of $1$, substituting $\theta =x/2$, and using $(2)$ reveals

$$\begin{align} \sin^2(x/2)&=\frac{1-\cos(x)}{2}\\\\ &\le x^2/4 \tag 3 \end{align}$$

whereupon we find for all $x$

$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$

For values of $x<2$, $\frac12 x^2<2$ and $(4)$ provides a tighter bound than $1-\cos(x)\le 2$. For $x\ge 2$, it is still correct that $1-\cos(x)\le \frac12 x^2$, but the inequality $1-\cos(x)\le 2$ is obviously tighter. Therefore, we can write

$$1-\cos(x)\le \begin{cases}\frac12 x^2&,x<2\\\\2&,x\ge 2\tag 5\end{cases}$$

So, $(5)$ provides a guideline for the appropriate use of the bounds for $1-\cos(x)$.

Solution 2:

Let $x\gt 0$. You know that $\sin(x) < x$ for such $x$. Integrating $$ 1 - \cos(x) =\int_0^x \sin(t)\, dt < \int_0^x t\,dt = {t^2\over 2}$$