Print cos(2π/17) exactly
CJam (60 bytes)

1-4{"(- +sqrt( * -4*
))/2"S/@a*N/\a*}:F~'-\+WF'-\+W-4FWFF'/2


Online demo

Output (with added newlines to avoid wrapping):

(--(--(-1+sqrt(1*1-4*-4))/2+sqrt(-(-1+sqrt(1*1-4*-4))/2*-(-1+sqrt(1*1-4*-4))/2-4*-1)
)/2+sqrt(-(--(-1+sqrt(1*1-4*-4))/2+sqrt(-(-1+sqrt(1*1-4*-4))/2*-(-1+sqrt(1*1-4*-4))/
2-4*-1))/2*-(--(-1+sqrt(1*1-4*-4))/2+sqrt(-(-1+sqrt(1*1-4*-4))/2*-(-1+sqrt(1*1-4*-4)
)/2-4*-1))/2-4*(-(--1+sqrt(-1*-1-4*-4))/2+sqrt((--1+sqrt(-1*-1-4*-4))/2*(--1+sqrt(-1
*-1-4*-4))/2-4*-1))/2))/2/2


I suspect this is the kind of answer OP was secretly hoping for. Rather than apply general-purpose compression to a string, it uses the field structure which Gauss first exploited to present this value in surds, in his Disquisitiones Arithmeticae.

I worked from this presentation (pages 18 and 19) but adapted it slightly in order to only ever require the positive root of a quadratic. Unrolled and in pseudo-code, the expression breaks down as

# Positive root of x^2 + bx + c = 0
F(b, c) = (-b + sqrt(b*b - 4*c))/2
# (8,1)
p = F(1, -4)
# (4,1)
q = F(-p, -1)
# -(8,3)
r = F(-1, -4)
# (4,3)
s = F(r, -1)
# Result
F(-q, s)/2