# Prove the following inequalities: $\sin x> x-\large\frac{x^{3}}{6}, $$x>0 Note: This is part 2nd of a 4 part question, split as 4 separate questions here. ## 1 Answer Toolbox: • (i) If f' is positive on an open interval I. Then f is strictly increasing on I • (ii) If f' is negative on an open interval I, then f is strictly decreasing on I • A function f(x) is said to be increasing on an interval I f(x_1) \leq f(x_2) whenever x_1 < x_2 • It is strictly increasing if x_1 < x_2 =>f(x_1) < f(x_2) • A function f is said to be decreasing on an interval I if f(x_1) \geq f(x_2) whenever x_1 < x_2 • It is strictly decreasing if x_1 < x_2 =>f(x_1) > f(x_2) \sin x > x -\large\frac{x^3}{6},$$ x > 0$
Let $f(x) =\sin x -(x-\large\frac{x^3}{6})$
Step 1:
$f'(x) =\cos x -1 +\large\frac {x^2}{2}$
$\qquad =\cos x -(1-\large\frac {x^2}{2})$
Step 2:
$f''(x)= -\sin x +x$
Step 3:
$f'''(x)=1-\cos x =2 \sin ^2 \large\frac{x}{2} $$> 0 for all x \therefore \;f''(x) is a strictly increasing function for x >0 f''(x) > f''(0) x- \sin x >0 for all x >0 \therefore \;f'(x) is a strictly increasing function for x >0 Step 4: f'(x) >f' (0) \quad x >0 =>\cos x -(1-\large\frac{x^2}{2})$$ > 1-(1-0)$
=>$\cos x- (1-\large\frac{x^2}{2}) $$> 0 =>f'(x) > 0 Step 5: \therefore \;f(x) is a strictly increasing function for x >0 =>f(x) > f(0) for x > 0 \therefore \sin x-(x -\large\frac{x^3}{6})$$> \sin 0-(0-\large\frac{0}{6}),$$x >0 \sin x -(x- \large\frac{x^3}{6})$$ > 0$ for , $x >0$
=> $\sin x > (x-\large\frac{x^3}{6})$ for $x > 0$