1. ## Inequality question

If $0< x <\frac{\pi}{2}$ prove

$sin(x) > x - \frac{x^{3}}{4}$

2. ## Re: Inequality question

Originally Posted by Mahan1
If $0< x <\frac{\pi}{2}$ prove

$sin(x) > x - \frac{x^{3}}{4}$
$\noindent Recall a continuous function f is said to be increasing if f'(x) > 0 (namely, it derivative is positive) for all values of x on some interval.$

$\noindent Now let f(x) = \sin x - x + \frac{x^3}{4}. So$

\begin{align*}f'(x) &= \cos x - 1 + \frac{3x^2}{4}\\f''(x) &= -\sin x + \frac{3x}{2}\\f'''(x) &= -\cos x + \frac{3}{2}.\end{align*}

$\noindent Thus it should be clear that f'''(x) > 0 as -1 \leqslant \cos x \leqslant 1 for all x. So f''(x) is increasing and is positive when x > 0 since f''(0) = 0. This implies f'(x) is increasing and is again positive when x > 0 since f'(0) = 0. But this in turn implies that f(x) is increasing and is positive when x > 0 since f(0) = 0.$

$\noindent Thus f(x) = \sin x - x + \frac{x^3}{4} > 0 or \sin x > x - \frac{x^3}{4} for all x > 0, as required to prove.$

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