TITLE: Prove that $1-\frac{\sin(x)}{x} < x^2/2$ for $x \in (0, \pi/2)$
QUESTION [5 upvotes]: How do we prove that 
$$
(2) \,\,\,1-\frac{\sin(x)}{x} < \frac{x^2}{2},\text{  for }x \in (0, \pi/2)
$$
This somehow follows from 
(1) $\sin(x) < x < \tan(x)$ which is true for $x \in (0, \pi/2)$
But I don't see how it follows. 
So I know (1) and I want to prove (2). 
I think I succeeded to prove that 
(3) $1-\frac{\sin(x)}{x} < x^2$
by using (1) and by using the fact that $a^2 < a$ for $a \in (0,1)$ 
Here is my proof:  
$1 - \frac{\sin(x)}{x} < 1 - \cos(x) < 1 - (\cos(x))^2 = (\sin(x))^2 < x^2 $
Is my proof correct? 
Also, how do I get the stronger inequality (2).

REPLY [2 votes]: We can make the inequality slightly stronger. Taylor's theorem for $\sin(x)$ gives you
$$1-\frac{\sin(x)}{x}=\cos(a)\frac{x^2}{6}$$
for some $a\in(0,x)$ and
$$1-\frac{\sin(x)}{x}=\cos(a)\frac{x^2}{6}\leq\frac{x^2}{6} $$
which is in fact true for all $x\in\mathbb{R}$.