Express cos(4x) in terms of powers of cos(x)

By quoting De Moivre's theorem, (r(cos(x) + isin(x)))ⁿ = rⁿ(cos(nx) + isin(nx)), we can realise that cos(4x) is a result of the real parts of (cos(x) + isin(x))⁴

Using binomial expansion: (where c = cos(x) and s = sin(x))

(cos(x) + isin(x))⁴ = c⁴ + 4c³is + 6c²(is)² + 4c(is)³ + (is)⁴

= c⁴ +4c³is - 6c²s² - 4cis³ + s⁴

The real parts of this are c⁴, -6c²s² and s⁴

Therefore, cos(4x) = c⁴ - 6c²s² + s⁴

s² = 1 - c²

6c²s² = 6c²(1-c²) = 6c² - 6c⁴

s⁴ = (1 - c²)² = 1 - 2c² + c⁴

By substituting these in:

cos(4x) = c⁴ - 6c² + 6c⁴ + 1 - 2c² + c⁴

= 8c⁴ - 8c² + 1