# If cos(pi/4)cos(pi/8)cos(pi/16)...cos(pi/(2^2006))=2^(-a)cos(pi/b) thsn find a and b?

Jul 24, 2018

Contd...........

#### Explanation:

We have,

$\cos \left(\frac{\pi}{2} ^ 2\right) \cos \left(\frac{\pi}{2} ^ 3\right) \cos \left(\frac{\pi}{2} ^ 4\right) \ldots \cos \left(\frac{\pi}{2} ^ 2006\right) = {2}^{-} a \cos \left(\frac{\pi}{b}\right)$.

Observe that, there are $2005$ terms on the L.H.S. of this eqn.

Multiplying both sides by

$\left\{2 \sin \left(\frac{\pi}{2} ^ 2\right)\right\} \left\{2 \sin \left(\frac{\pi}{2} ^ 3\right)\right\} \ldots \left\{2 \sin \left(\frac{\pi}{2} ^ 2006\right)\right\}$, we get,

$\left\{2 \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}\right)\right\} \left\{2 \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right)\right\} \ldots \left\{2 \cos \left(\frac{\pi}{2} ^ 2006\right) \sin \left(\frac{\pi}{2} ^ 2006\right)\right\}$

$= \left({2}^{2005} \cdot {2}^{-} a\right) \cos \left(\frac{\pi}{b}\right) \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{8}\right) \ldots \sin \left(\frac{\pi}{2} ^ 2006\right)$.

$\therefore \sin \left(2 \cdot \frac{\pi}{4}\right) \sin \left(2 \cdot \frac{\pi}{8}\right) \ldots \sin \left(2 \cdot \frac{\pi}{2} ^ 2006\right) = {2}^{2005 - a} \cos \left(\frac{\pi}{b}\right) \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{8}\right) \ldots \sin \left(\frac{\pi}{2} ^ 2006\right) , i . e . ,$

$\sin \left(\frac{\pi}{2}\right) \cancel{\sin} \left(\frac{\pi}{4}\right) \ldots \cancel{\sin} \left(\frac{\pi}{2} ^ 2005\right) = {2}^{2005 - a} \cos \left(\frac{\pi}{b}\right) \cancel{\sin} \left(\frac{\pi}{4}\right) \ldots \cancel{\sin} \left(\frac{\pi}{2} ^ \left(2005\right)\right) \sin \left(\frac{\pi}{2} ^ 2006\right)$.

$\therefore 1 = {2}^{2005 - a} \cos \left(\frac{\pi}{b}\right) \sin \left(\frac{\pi}{2} ^ 2006\right)$.