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)#.
