A more fun way
By Bernoulli expansion,
(1-itanA))^8 = (1)^8 + 8(1)^7(-itanA) + 28(1)^6(-itanA)^2 + 56(1)^5(-itanA)^3 + 70(1)^4(-itanA)^4 + 56(1)^3(-itanA)^5 + 28(1)^2(-itanA)^6 + 8(1)(-itanA)^7 + (-itanA)^8
and
(1+ itan@)^8 = 2sec^8@cos8@ - (1-itan@)^8
Expanding (1-itan@)^8 in Bernoulli, real parts only occur when we have even powers of i
RE{1-itan@)^8 = - {1 - 28tan^2@ + 70tan^4@ - 28tan^6@ + tan^8@}
= - {1 -28(3 - 2sqrt(2)) + 70(17 - 12sqrt(2)) - 28(99 - 70sqrt(2)) + 577 - 408sqrt(2)}
= - {[-83+1190-2772+577] + [-56-840+1960-408]sqrt2)}
= 1088 - (768*sqrt2)
= 1088 - 1086.116015902536997479696940193
using
tan(pi/8) = sqrt(2) - 1
tan^2(pi/8) = 3 - 2sqrt(2)
tan^4(pi/8) = 17 - 12sqrt(2)
tan^6(pi/8) = 99 - 70sqrt(2)
tan^8(pi/8) = 577 - 408sqrt(2)
and taking real parts,
RE{(1+ itan@)^8} = 2sec^8@cos8@ + RE{ - (1-itan@)^8}
= -2{4-2sqrt(2)}^4 + 1088 - (768*sqrt2) = 1.8839840974631
= 64(12*sqrt2 - 17)
using sec(pi/8) = sqrt{4-2sqrt(2)}
sec^8(pi/8) = {4-2sqrt(2)}^4
cos(pi) = -1
QED