Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that

$$1- \frac{x}{3} \le \frac{\sin x}x \le 1.1 - \frac{x}{4}, \quad \forall x\in(0,\pi].$$

share|improve this question
add comment

1 Answer

From the concavity of $f(x)=\cos x$ over $[0,\pi/2]$, we have:

$$ \forall x\in[0,\pi/2],\quad \cos x\geq 1-\frac{2}{\pi}x, $$

from which

$$ \forall x\in[0,\pi/2],\quad \sin x\geq \frac{1}{\pi}x(\pi-x) = x-\frac{1}{\pi}x^2$$

follows, by integration. Now, both the RHS and the LHS are symmetric wrt $x=\frac{\pi}{2}$, so we can extend the inequality over the whole $[0,\pi]$ interval:

$$ \forall x\in[0,\pi],\quad \frac{\sin x}{x}\geq 1-\frac{x}{\pi}\geq 1-\frac{x}{3}, $$

proving the "easier" inequality. Using an analogue tecnique, it's possible to establish something a little weaker than the other desired inequality. Since:

$$\operatorname{argmax}_{[0,\pi]}\left(\cos x+\frac{x}{2}\right)=\frac{\pi}{6},$$

we have:

$$ \forall x\in[0,\pi],\quad \cos x+\frac{1}{2}-\left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)\leq 0,$$

so, by integration,

$$\forall x\in[0,\pi],\quad \sin x +\frac{x^2}{4} - \left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)x \leq 0,$$

holds, that is:

$$\forall x\in[0,\pi],\quad \frac{\sin x}{x}\leq 1.1278247\ldots-\frac{x}{4}.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.