Re: Show 1-x/3<sinx/x<1.1-x/4
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($$1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$) is to multiply it through by $x$ and hence what we need to prove now is this: $$x-\frac{x^2}{3}<\sin x<1.1x-\frac{x^2}{4}$$.[TABLE="class: grid, width: 300"]
[TR]
[TD]First, we consider the part to prove $x-\frac{x^2}{3}<\sin x$.[/TD]
[TD]Next, we need to prove $$\sin x< 1.1x-\frac{x^2}{4}$$.
[/TD]
[/TR]
[TR]
[TD]Let
$$f(x)=\sin x-x+\frac{x^2}{3}$$
Differentiating it once we get:
$$f'(x)=\cos x-1+\frac{2x}{3}$$
and differentiating it again we have:
$$f''(x)=-\sin x+\frac{2}{3}$$[/TD]
[TD]Let
$$g(x)=\sin x-1.1x+\frac{x^2}{4}$$
Differentiating it once we get:
$$g'(x)=\cos x-1.1+\frac{x}{2}$$
and differentiating it again we have:
$$g''(x)=-\sin x+\frac{1}{2}$$[/TD]
[/TR]
[TR]
[TD]Observe that $$f"(x)>0$$ for $$0<x< \pi$$ except when $$a<x<b$$ where $$\sin a= \sin b=\frac{2}{3}$$ and this implies also $$a<\frac{\pi}{2}<b$$.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 $$f'(b)=\cos b-1+\frac{2b}{3}>\cos (\frac{\pi}{2})-1+\frac{2(\frac{\pi}{2})}{3}>\frac{\pi}{3}-1>0$$.
From this we know that $f'(x)\ge 0$ for $0 \le x \le \pi$ and consequently, $f(x)$ increases from $0$ to $$\pi(\frac{\pi}{3}-1)$$ and this gives $$\sin x\ge x-\frac{x^2}{3}$$.[/TD]
[TD]By using the similar concept that applied in the previous
case, we have
$g"(x) \ge 0$ except for $$\frac{\pi}{6}<x<\frac{5\pi}{6}$$ where $g"(x) < 0$. So $g'(x)$ increases from a value of $-1.1$ to a maximum of $$\cos \frac{\pi}{6}-1.1+\frac{\frac{\pi}{6}}{2}>0$$. It must be zero at a point, $$x=p (0\le p \le \frac{\pi}{6}$$.
Similarly, $g'(q)=0$ for $$\frac{\pi}{6}<x<\frac{5\pi}{6}$$. Note that $$g'(\frac{\pi}{4})=\frac{\sqrt{2}}{2}-1.1+\frac{3\pi}{8}<0$$, so $$ \frac{\pi}{4}\le q \le \frac{3\pi}{4}$$. 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 $0\le x\le \pi$.[/TD]
[/TR]
[/TABLE]
And therefore we prove that $$1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$ is true for $0<x\le \pi$.
