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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.

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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$
answered Jul 30, 2013 by meena.p
 
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