Inequality question (1 Viewer)

Mahan1 · Feb 19, 2017

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

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

omegadot · Feb 24, 2017

Mahan1 said:
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.$

Inequality question (1 Viewer)

Mahan1

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omegadot

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