MHB Show 1 - x/3 < sinx/x < 1.1 - x/4

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion focuses on proving the inequality \(1 - \frac{x}{3} < \frac{\sin x}{x} < 1.1 - \frac{x}{4}\) for \(0 < x \leq \pi\). The proof involves manipulating the inequality by multiplying through by \(x\) to establish two separate inequalities: \(x - \frac{x^2}{3} < \sin x\) and \(\sin x < 1.1x - \frac{x^2}{4}\). The function \(f(x) = \sin x - x + \frac{x^2}{3}\) is shown to be non-decreasing, confirming the left side of the inequality, while the function \(g(x) = \sin x - 1.1x + \frac{x^2}{4}\) is analyzed to demonstrate that it remains non-positive, thus validating the right side. Consequently, the original inequality is proven true for the specified range of \(x\).
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Show that $$1-\frac{x}{3}<\frac{\sin x}{x}<1.1-\frac{x}{4}$$ for $0<x\le \pi$.
 
Mathematics news on Phys.org
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 $0<x\le \pi$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top