It is easy to derIve the irrational expression of sin18:

Starting with identity
 sin 72° = 2 sin 36° cos 36°                                 thru
cos 18° = 2 sin 36° cos 36°                               
  we arrive to quadratic equation
with one relevant root
t= 1/4*(sqrt5-1)                                                       and    t^2=1/8* (-sqrt5+3)         
where            t=  sin 18°

Now, back to our problem:

Applying few trigonmetric trivial identities ,we get

Z=sin 54º-sin 18=sin18º*(2*(cos18º)^2t*cos36º-1)=
=t*(2-2t^2+1)

Z=1/4*(sqrt5-1)*(2-4*1/8* (-sqrt5+3) )=
2*Z=(sqrt5-1)*(sqrt5+1)/4=(5-1) /4=1

Z=1/2   QED