Calculating cos(11pi/8)

How am I supposed to separate 11pi/8 into a mixture of pi/2, pi/3, pi/4 and pi/6? Can't any combination that achieves it.

11pi/8 = (11pi/4) / 2

t = 11pi/8

cos(2t) =>
cos(t)^2 - sin(t)^2 =>
cos(t)^2 - (1 - cos(t)^2) =>
2cos(t)^2 - 1

2cos(11pi/8)^2 - 1 = cos(11pi/4)
2 * cos(11pi/8)^2 - 1 = sqrt(2)/2
2 * cos(11pi/8)^2 = (2 + sqrt(2)) / 2
cos(11pi/8)^2 = (2 + sqrt(2)) / 4
cos(11pi/8) = +/- sqrt(2 + sqrt(2)) / 2

11pi/8 is in Q3, so cos(11pi/8) is < 0

-sqrt(2 + sqrt(2)) / 2

cos(11π/8)
-sin(π/8)