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.
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