1. ## Very hard inequality

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

2. Originally Posted by roshanhero
Use the inequalities $\displaystyle \sin \theta < \theta < \tan \theta$ for a suitable value of $\displaystyle \theta$ to show that $\displaystyle \pi$ lies between 3 and $\displaystyle 2\sqrt{3}$.
Let $\displaystyle \theta = \sin^{-1}x$ then $\displaystyle x < \sin^{-1} x < \frac{x}{\sqrt{1-x^2}}$ and then choose $\displaystyle x = \frac{1}{2}$ .

3. Hello, roshanhero!

A fascinating problem . . .

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

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

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

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

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

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

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

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

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

4. What about using the calculator ???

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

6. Originally Posted by roshanhero
How can we show that the inequality $\displaystyle \sin \theta < \theta$ holds for all values of $\displaystyle \theta$ greater than 0.
There are many ways. For example, the assertion is clear if $\displaystyle \theta\geqslant 1$ so assume $\displaystyle 0<\theta<1$. We know that $\displaystyle \sin(\theta)=\theta-\frac{\theta^3}{3!}+\cdots$ using some clever manipulation of the series, and using the fact that $\displaystyle 0<\theta<1\implies \theta^{n\geqslant 1}\leqslant \theta$ we may conclude that $\displaystyle \sin(\theta)=\theta+\text{negative terms}$.

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

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

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