Welcome to our community

Be a part of something great, join today!

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

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Show that \(\displaystyle 1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}\) for $0<x\le \pi$.
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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 \(\displaystyle 0<x< \pi\) 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 $0<x\le \pi$.