Another question:

Given that the binomial expansion of (z + (1/z)) ^ 5 = z^5 + 5z^3 + 10z + 10z^-1 + 5z^-3 + z^-5 = (2 cos n theta)^5

Show that cos^5 theta = 1/16 (a cos 5 theta + b cos 3 theta + c cos theta)