` Cos { 2 pi/2^64 - 1) cos{2^2 pi/(2^64-1)}.........cos{2^64 pi/(2^64 - 1)}` =

cos{(2pi)/(2^(64)-1)}cos{(2^2pi)/(2^(64)-1)}......cos{(2^(64)pi)/(2^(64)-1)}=

cos {2 (pi) / (2 ^ (64)) - 1) cos {2 ^ (2) (pi) / (2 ^ (64) -1)} dots ... cos {2 ^ (64) ( pi) / (2 ^ (64) -1)}

the value of cos((pi)/(2^2)) cos((pi)/(2^3)) cos((pi)/(2^4)) ..... cos((pi)/(2^(10))) sin((pi)/(2^(10))) is (a) (1)/(1024) (b) (1)/(512) (c) (1)/(256) (d) (1)/(128)

sin(pi/2^2009)cos(pi/2^2009)cos(pi/2^2008).....cos(pi/2^2) is

sin(pi/2^2009)cos(pi/2^2009)cos(pi/2^2008).....cos(pi/2^2) is

sin ((pi) / (2 ^ (2009))) cos ((pi) / (2 ^ (2009))) cos ((pi) / (2 ^ (2008))) ...... cos ( (pi) / (2 ^ (2)))

cos ((pi) / (65)) cos ((2 pi) / (65)) cos ((4 pi) / (65)) cos ((8 pi) / (65)) cos ((16 pi) / (65)) cos ((32 pi) / (64)) = (1) / (64)

(sin ^ (2) [64]) / (cos ^ (2) [26]) =?

prove that 4(cos(2 pi))/(7)*(cos pi)/(7)-1=2(cos(2 pi))/(7)

If quad (cos^(2)pi)/(2)+(cos^(2)(2 pi))/(n)+...+(cos^(2)((n-1)pi))/(n) then s equals