# Usecos(x/2) * cos(x/4) * cos(x/8) and the half angle formula [cos(A/2) = sqrt(1+cosA)/2] to show 2/π = sqrt(2)/2 * (sqrt(2+sqrt2))/2 ... How?

Sep 16, 2016

See the Proof in Explanation.

#### Explanation:

We will use the following Identities :=

$\left(1\right) : 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right) = \sin \theta .$

$\left(2\right) : \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} = {\left(\frac{1 + \cos \theta}{2}\right)}^{\frac{1}{2}} .$

$\left(3\right) : {\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$.

The Exp.$= \cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right)$.

In the First Step, we multiply by $2 \sin \left(\frac{x}{2}\right)$ in Nr. & Dr., to get,

The Exp.={(2cos(x/2)sin(x/2))cos(x/4)cos(x/8)}/(2sin(x/2)

$= \frac{\sin x \cos \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right)}{2 \sin \left(\frac{x}{2}\right)} \ldots \ldots \ldots \ldots \ldots . \left[b y , \left(1\right)\right]$

We continue in the same fashion, and, find that,

The Exp.=((sinx)(2sin(x/4)cos(x/4))cos(x/8))/((2sin(x/2)2sin(x/4))

$= \frac{\sin x \cancel{\sin} \left(\frac{x}{2}\right) \cos \left(\frac{x}{8}\right)}{4 \cancel{\sin} \left(\frac{x}{2}\right) \sin \left(\frac{x}{4}\right)}$

=(sinxcos(x/8))/(4sin(x/4)

$= \frac{\sin x \left(2 \sin \left(\frac{x}{8}\right) \cos \left(\frac{x}{8}\right)\right)}{8 \sin \left(\frac{x}{4}\right) \sin \left(\frac{x}{8}\right)}$

$= \frac{\sin x \cancel{\sin} \left(\frac{x}{4}\right)}{8 \cancel{\sin} \left(\frac{x}{4}\right) \sin \left(\frac{x}{8}\right)}$

$= \sin \frac{x}{8 \sin \left(\frac{x}{8}\right)}$

Thus, we have proved that,

$\cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{4}\right) \cos \left(\frac{x}{8}\right) = \sin \frac{x}{8 \sin \left(\frac{x}{8}\right)}$

Generalising,

cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^n)=sinx/(2^nsin(x/2^n)

$= \left\{\frac{\sin \frac{x}{x}}{\sin \frac{\frac{x}{2} ^ n}{\frac{x}{2} ^ n}}\right\} \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$

Now, as $n \rightarrow \infty , {2}^{n} \rightarrow \infty , \frac{x}{2} ^ n \rightarrow 0 , \sin \frac{\frac{x}{2} ^ n}{\frac{x}{2} ^ n} \rightarrow 1$.

$\therefore \text{, by } \left(\star\right) ,$

$\cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{2} ^ 2\right) \cos \left(\frac{x}{2} ^ 3\right) \ldots \cos \left(\frac{x}{2} ^ n\right) \ldots \ldots = \sin \frac{x}{x}$

To get the desired result, let us take, $x = \frac{\pi}{2}$ in this last eqn.,

$\cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{16}\right) \ldots \ldots . = \sin \frac{\frac{\pi}{2}}{\frac{\pi}{2}} = \frac{2}{\pi}$, whereas,

On the. L.H.S., $\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} ,$

$\cos \left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Therefore,

$\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \ldots \ldots \ldots \ldots = \frac{2}{\pi}$

Enjoy maths.!