# Prove the following inequalities: $\cos x> 1-\large\frac{x^{2}}{2}, $$x>0 Note: This is part 1st of a 4 part question, split as 4 separate questions here. ## 1 Answer Toolbox: • (i) If f' is positive on an open interval I. Then f is strictly increasing on I • (ii) If f' is negative on an open interval I, then f is strictly decreasing on I • A function f(x) is said to be increasing on an interval I f(x_1) \leq f(x_2) whenever x_1 < x_2 • It is strictly increasing if x_1 < x_2 =>f(x_1) < f(x_2) • A function f is said to be decreasing on an interval I if f(x_1) \geq f(x_2) whenever x_1 < x_2 • It is strictly decreasing if x_1 < x_2 =>f(x_1) > f(x_2) \cos x > 1- \large\frac{x^2}{2}$$, x >0$
Step 1:
Let $f(x)=\cos x -\bigg(1-\large\frac{x^2}{2}\bigg), $$x >0 f'(x)=\sin x +x Step 2: f''(x)=1- \cos x= 2 \sin ^2 \large\frac{x}{2}$$ >0$ for all x
$\therefore \;f'(x)$ is a strictly increasing function for $x >0$
$\therefore f'(x) > f'(0) \;for\; x>0$
$=> x -\sin x >0$
Step 3:
We have $f'(x)=x -\sin x>0$ for $x>0$
$\therefore f(x)$ is a strictly increasing function
Step 4:
=>$f(x) >f(0) \;for\; x >0$
$\cos x -\bigg(1-\large\frac{x^2}{2}\bigg)$$>1-(1-0) \cos x -\bigg(1-\large\frac{x^2}{2}\bigg)$$ >0$
$\therefore \cos x > 1-\large\frac{x^2}{2}$ for $x>0$