Correct Answer - C
We have that,
`cos alpha. cos(2alpha) cos(2^(2)alpha)...cos(2^(n-1)alpha)=(sin(2^(n)alpha))/(2^(n)sinalpha)`
`therefore cos .(pi)/(2^(2)). cos. (pi)/(2^(3)) ... cos.(pi)/(2^(10)).sin.(pi)/(2^(10))`
`={(sin((pi)/(2^(10))2^(9)))/(2^(9)sin((pi)/(2^(10))))}sin .(pi)/(2^(10))" "[because "here", alpha = (pi)/(2^(10)) and n = 9]`
`=(1)/(2^(9))sin((pi)/(2)) = (1)/(2^(9))=(1)/(512)`
