Very hard inequality

**roshanhero** · February 10th 2010, 09:10 AM

Use the inequalities $\sin \theta < \theta < \tan \theta$ for a suitable value of $\theta$ to show that $\pi$ lies between 3 and $2\sqrt{3}$ .

**Danny** · February 10th 2010, 10:22 AM

Originally Posted by roshanhero

Use the inequalities $\sin \theta < \theta < \tan \theta$ for a suitable value of $\theta$ to show that $\pi$ lies between 3 and $2\sqrt{3}$ .

Let $\theta = \sin^{-1}x$ then $x < \sin^{-1} x < \frac{x}{\sqrt{1-x^2}}$ and then choose $x = \frac{1}{2}$ .

**Soroban** · February 10th 2010, 10:23 AM

Hello, roshanhero!

A fascinating problem . . .

Use the inequalities: $\sin \theta \,<\, \theta \,<\, \tan \theta$ for a suitable value of $\theta$

to show that $\pi$ lies between 3 and $2\sqrt{3}$ .

We have: . $\theta \:>\:\sin\theta$
$\sin\theta$ is an increasing function on the interval $\left[0,\:\frac{\pi}{2}\right]$
. . Then: . $\arcsin\theta \:>\:\theta$

Let $\theta \,=\,\frac{1}{2}$
We have: . $\arcsin\left(\frac{1}{2}\right) \:>\:\frac{1}{2}\quad\Rightarrow\quad \frac{\pi}{6} \:>\:\frac{1}{2}$

. . Hence: . ${\color{blue}\pi \:>\:3}$

We have: . $\theta \:<\:\tan\theta$
$\tan\theta$ is an increasing function on the interval $\left[0,\:\frac{\pi}{2}\right]$
. . Then: . $\arctan\theta \:<\:\theta$

Let $\theta = \frac{1}{\sqrt{3}}$
We have: . $\arctan\left(\frac{1}{\sqrt{3}}\right) \:<\:\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \frac{\pi}{6} \:<\:\frac{1}{\sqrt{3}}$

. . Hence: . $\pi \:<\:\frac{6}{\sqrt{3}} \quad\Rightarrow\quad {\color{blue}\pi \:<\:2\sqrt{3}}$

Therefore: . $\boxed{\;3 \;<\;\pi \;<\;2\sqrt{3}\;}$

**TWiX** · February 10th 2010, 10:26 AM

What about using the calculator ???

**roshanhero** · February 10th 2010, 10:49 AM

How can we show that the inequality $\sin \theta < \theta$ holds for all values of $\theta$ greater than 0.

**Drexel28** · February 10th 2010, 11:12 AM

Originally Posted by roshanhero

How can we show that the inequality $\sin \theta < \theta$ holds for all values of $\theta$ greater than 0.

There are many ways. For example, the assertion is clear if $\theta\geqslant 1$ so assume $0<\theta<1$ . We know that $\sin(\theta)=\theta-\frac{\theta^3}{3!}+\cdots$ using some clever manipulation of the series, and using the fact that $0<\theta<1\implies \theta^{n\geqslant 1}\leqslant \theta$ we may conclude that $\sin(\theta)=\theta+\text{negative terms}$ .

Alternatively, the claim that $\sin(\theta)<\theta$ is equivalent to $0<\theta-\sin(\theta)=\theta-\sin(\theta)-(0-\sin(0))=\int_0^{\theta}\left(1-\cos(x)\right)dx$ . Since an integral is positive precisely when it's integrand is positive over the region of integration all we need to show is that $0\leqslant x<\theta<1\implies 1-\cos(x)\geqslant 0$ but, this follows readilly.

Lastly, the easiest way is to just note that if $f(\theta)=\theta-\sin(\theta)$ then $f(0)=0,f'(\theta)=1-\cos(\theta)\geqslant 0$ and so $f(\theta)\geqslant 0,\text{ }\theta\geqslant 0$ .

Or, actually. You could use the fact that $f(x)=\sin(x)$ is Lipschitz (with Lipschitz constant one) to conclude that since $\left|\sin(x)-\sin(y)\right|\leqslant |x-y|,\text{ }\forall x,y\in\mathbb{R}\implies |\sin(x)-\sin(0)|=|\sin(x)|\leqslant |x|$ . Which, is a stronger claim.

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