Let eta = [3(cos 14^@ + isin 14^@)]^(-4)
We can see it resembles the trigonometric form of complex numbers :
z = [r(costheta + isintheta)]^(t)
We can "mess" around with this identity to furthermore get
z = r^t[cos(t*theta) + isin(t*theta)], derived from de Moivre's theorem.
In our case, r = 3, theta = 14^@ and t = -4.
eta = 1/(3^4)(cos(-56^@) + isin(-56^@)).
By parity of sin and cos functions, we have
eta = 1/(3^4)(cos(56^@) - isin(56^@)).
Right now, it would've been much nicer if we had theta = 15^@, but we'll try to do it anyway.
Both sin(56^@) and cos(56^@) have "ugly" forms when written in exact value. The best way to do this would be to approximate them with sin(60^@) and cos(60^@), respectively.
When doing so, you'll find out that sin(56^@) differs from sin(60^@) only by about ~ -0.03 and cos(56^@) by ~0.05 from cos(60^@).
Plugging these values into our equation, we have :
eta ~~1/(3^4)(cos(60^@) + 0.05 - i(sin(60^@)-0.03))
eta ~~ 1/(3^4)(1/2 + 1/20 - i(sqrt3/2 - 3/100))
eta ~~ 11/1620 - i[1/(27sqrt3) - 1/2700]
