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- Feb 14, 2012

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Show that \(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}\) for $0<x\le \pi$.

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- Feb 14, 2012

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Show that \(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}\) for $0<x\le \pi$.

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- Feb 14, 2012

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Hi MHB,

The solution presented below is proposed by Georges Ghosn:

Since the inequality defined for $0<x\le \pi$, what we could do with the given inequality(\(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}\)) is to multiply it through by $x$ and hence what we need to prove now is this: \(\displaystyle x-\frac{x^2}{3}<\sin x<1.1x-\frac{x^2}{4}\).

First, we consider the part to prove $x-\frac{x^2}{3}<\sin x$. | Next, we need to prove \(\displaystyle \sin x< 1.1x-\frac{x^2}{4}\). |

Let \(\displaystyle f(x)=\sin x-x+\frac{x^2}{3}\) Differentiating it once we get: \(\displaystyle f'(x)=\cos x-1+\frac{2x}{3}\) and differentiating it again we have: \(\displaystyle f''(x)=-\sin x+\frac{2}{3}\) | Let \(\displaystyle g(x)=\sin x-1.1x+\frac{x^2}{4}\) Differentiating it once we get: \(\displaystyle g'(x)=\cos x-1.1+\frac{x}{2}\) and differentiating it again we have: \(\displaystyle g''(x)=-\sin x+\frac{1}{2}\) |

Observe that \(\displaystyle f"(x)>0\) for \(\displaystyle 0<x< \pi\) except when \(\displaystyle a<x<b\) where \(\displaystyle \sin a= \sin b=\frac{2}{3}\) and this implies also \(\displaystyle a<\frac{\pi}{2}<b\). Thus, $f'(x)$ increases from $0$ to a maximum at $x=a$, then decreases to a minimum at $x=b$, and then increases again from $b$ to $\pi$. The minimum value is \(\displaystyle f'(b)=\cos b-1+\frac{2b}{3}>\cos (\frac{\pi}{2})-1+\frac{2(\frac{\pi}{2})}{3}>\frac{\pi}{3}-1>0\). From this we know that $f'(x)\ge 0$ for $0 \le x \le \pi$ and consequently, $f(x)$ increases from $0$ to \(\displaystyle \pi(\frac{\pi}{3}-1)\) and this gives \(\displaystyle \sin x\ge x-\frac{x^2}{3}\). | By using the similar concept that applied in the previous case, we have $g"(x) \ge 0$ except for \(\displaystyle \frac{\pi}{6}<x<\frac{5\pi}{6}\) where $g"(x) < 0$. So $g'(x)$ increases from a value of $-1.1$ to a maximum of \(\displaystyle \cos \frac{\pi}{6}-1.1+\frac{\frac{\pi}{6}}{2}>0\). It must be zero at a point, \(\displaystyle x=p (0\le p \le \frac{\pi}{6}\). Similarly, $g'(q)=0$ for \(\displaystyle \frac{\pi}{6}<x<\frac{5\pi}{6}\). Note that \(\displaystyle g'(\frac{\pi}{4})=\frac{\sqrt{2}}{2}-1.1+\frac{3\pi}{8}<0\), so \(\displaystyle \frac{\pi}{4}\le q \le \frac{3\pi}{4}\). Then from $0$, $g(x)$ decreases to a minimum at $x=p$, then increases to a maximum value at $x=q$. $g'(q)=0=\cos q-1.1+\frac{q}{2}$ or $q=2(1.1-\cos q)$. Thus, $g(q)=\sin q-1.1q+\frac{q^2}{4}=\sin q+\cos^2 q-(1.1)^2=(0.03-\sin q)(\sin q-0.7)<0$ since $\sin q \ge \frac{\sqrt{2}}{2}$. From this $g(x)=\sin x-1.1+\frac{x^2}{4} \le 0$ for $0\le x\le \pi$. |

And therefore we prove that \(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}\) is true for $0<x\le \pi$.