Find cos4x in terms of cosx.

As with any question involving sines and cosines, consider complex numbers as a likely way to find the answer. For this particular kind of question, where sin/cos of a multiple of x is needed in terms of sinx/cosx, or vice versa, the complex number representation z = cosx + isinx must be expanded in two ways. 

Firstly, expand using de Moivre's Theorem:

z = cosx + isinx

z⁴ = (cosx + isinx)⁴ =  cos4x + sin4x

Then, expand using the binomial expansion formula to get powers of cosx:

z⁴ = (cosx + isinx)⁴ = cos⁴x + 4icos³xsinx + 6i²cos²xsin²x + 4i³cosxsin³x + i⁴sin⁴x

                               = cos⁴x + 4icos³xsinx - 6cos²xsin²x - 4icosxsin³x + sin⁴x

Equate the real parts of both expansions to get cos equivalence:

cos4x = cos⁴x - 6cos²xsin²x + sin⁴x

Use cos²x + sin²x = 1 as a substitution:

cos4x = cos⁴x - 6cos²x(1-cos²x) + (1-cos²x)²

          = 8cos⁴x - 8cos²x + 1