Rearrange cos^3x... (1 Viewer)

gr_111 · Jun 8, 2012

G'day

Can someone show me the steps to prove (cos³x) = (1/4(3cosx + cos3x))???
Thanks.

nightweaver066 · Jun 8, 2012

gr_111 said:
G'day

Can someone show me the steps to prove (cos³x) = (1/4(3cosx + cos3x))???
Thanks.

$cos3x = cos(x + 2x)\\=cosxcos2x - sinxsin2x\\=cosx(2cos^2x -1) - 2sin^2xcosx\\=2cos^3x - cosx - 2cosx(1 - cos^2x)\\=2cos^3x - cosx - 2cosx + 2cos^3x\\=4cos^3x - 3cosx\\~\therefore cos3x = 4cos^3x - 3cosx\\$Rearranging in terms of $ cos^3x,\\4cos^3x = cos3x + 3cosx\\cos^3x = \frac{1}{4}(cos3x + 3cosx)$

gr_111 · Jun 8, 2012

Ok, so you don't actually start with cos³x, but rather cos3x...?

gr_111 · Jun 8, 2012

Thanks for that, rep'd.

Rearrange cos^3x... (1 Viewer)

gr_111

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nightweaver066

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