# Show 1 - x/3 < sinx/x < 1.1 - x/4

#### anemone

##### MHB POTW Director
Staff member
Show that $$\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$ for $0<x\le \pi$.

#### anemone

##### MHB POTW Director
Staff member
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($$\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$) 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 $$\displaystyle 0\cos (\frac{\pi}{2})-1+\frac{2(\frac{\pi}{2})}{3}>\frac{\pi}{3}-1>0$$. 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 $$\displaystyle \frac{\pi}{6}0$$. 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} And therefore we prove that \(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$ is true for $0<x\le \pi$.