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

1 Answer
Jul 24, 2018

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

Explanation:

We have,

#cos(pi/2^2)cos(pi/2^3)cos(pi/2^4)...cos(pi/2^2006)=2^-acos(pi/b)#.

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

Multiplying both sides by

#{2sin(pi/2^2)}{2sin(pi/2^3)}...{2sin(pi/2^2006)}#, we get,

#{2cos(pi/4)sin(pi/4)}{2cos(pi/8)sin(pi/8)}...{2cos(pi/2^2006)sin(pi/2^2006)}#

#=(2^2005*2^-a)cos(pi/b)sin(pi/4)sin(pi/8)...sin(pi/2^2006)#.

#:. sin(2*pi/4)sin(2*pi/8)...sin(2*pi/2^2006)=2^(2005-a)cos(pi/b)sin(pi/4)sin(pi/8)...sin(pi/2^2006), i.e., #

#sin(pi/2)cancelsin(pi/4)...cancelsin(pi/2^2005)=2^(2005-a)cos(pi/b)cancelsin(pi/4)...cancelsin(pi/2^(2005))sin(pi/2^2006)#.

#:. 1=2^(2005-a)cos(pi/b)sin(pi/2^2006)#.