Express cos5x in terms of increasing powers of cosx

De Moivre's theorem: (cos5x + isin5x) = (cosx + isinx)⁵ To get cos5x we will need to expand (cosx + isinx)⁵ and then take the real parts. Binomial expansion: (cosx + isinx)⁵ = (5C0)(cosx)⁵ + (5C1)(cosx)⁴(isinx) + (5C2)(cosx)³(isinx)² + (5C3)(cosx)²(isinx)³ + (5C4)(cosx)(isinx)⁴ + (5C5)(isinx)⁵ = cos⁵x + 5i(cos⁴x)(sinx) - 10(cos³x)(sin²x) - 10i(cos²x)(sin³x) + 5(cosx)(sin⁴x) + isin⁵x Taking the real parts of the above expansion to get cos5x: cos5x = cos⁵x - 10(cos³x)(sin²x) + 5(cosx)(sin⁴x) To get cos5x in terms of powers of cosx only, we need to use the trigonometric identity cos²x + sin²x = 1 Rearrange this to get sin²x = 1 - cos²x and substitute in: cos5x = cos⁵x - 10(cos³x)(1 - cos²x) + 5(cosx)(1 - cos²x)² And expand: = cos⁵x - 10(cos³x)(1 - cos²x) + 5(cosx)(cos⁴x - 2cos²x + 1) = cos⁵x - 10cos³x + 10cos⁵x + 5cos⁵x - 10cos³x + 5cosx Finally group like powers together: cos5x = 5cosx - 20cos³x +16cos⁵x In order to check this is correct, substitute in different values of x for your answer and cos5x.