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
LD
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