Showing that $1-\cos(x) <\frac{x^2}{2}$ holds

Question

Show that for all $x \in (0, \frac\pi2)$ the inequality $1-\cos(x) <\frac{x^2}{2}$ holds.

Setting $f(x) = \frac{x^2}{2}+\cos(x)-1$ and differentiating it results in $f'(x)=x-\sin(x)$.

Differentiating the function once more we get $f''(x)=1-\cos(x) \geqslant 0$, since $-1 \leqslant \cos(x) \leqslant 1$.

So we know that $f'$ is strictly increasing. My question is, how should we continue from here?

Answer 1

Since $f'$ is increasing, we have $0=f'(0) \leq f'(x)$ for any$x\geq0$

Answer 2

Note that $1>\cos u $ for $x\in (0, \frac\pi2)$. Then,

$$\int_0^x \int_0^t (1-\cos u)du dt=\frac{x^2}{2}-(1-\cos x) > 0$$

Thus,

$$1-\cos x<\frac{x^2}{2}$$

Answer 3

Using $$ |\sin(x)|\le |x| $$ one has $$ 1-\cos(x)=2\sin^2(\frac x2)\le 2|\frac{x}{2}|^2=\frac{x^2}{2}.$$