96`sqrt3 sin(pi/48) cos(pi/48) cos(pi/24) cos(pi/12) cos(pi/6)` has the value

Explanation

Here,
`2^(n-1) = (pi//6)/(pi//48)=8`
`2^(n-1) = 2^(3)`
n=4
Applying,
`cos alpha.cos2alpha………… cos2^(n-1) alpha = (sin2^(n)alpha)/(2^(n)sinalpha)`
`=96sqrt(3) xx sin pi/48 xx (sin2^(4) xx pi/48)/(2^(4) sinpi/48)`
`=(96sqrt(3))/(16) xx sin pi/3`
`=6 xx sqrt(3) xx sqrt(3)/2`
=9