Asked by Rin
By expressing 3x as (2x + x) prove that
cos3x = 4cos^3x - 3cosx
cos3x = 4cos^3x - 3cosx
Answers
Answered by
Reiny
cos(3x)
= cos(2x + x)
= cos2xcosx - sin2xsinx
= (2cos^2 x -1)cosx - (2sinxcosx)sinx
= 2cos^3 x - cosx - 2sin^2 x (cosx)
= 2 cos^3x - cosx - 2(1 - cos^2 x)(cosx)
= 2 cos^3x - cosx - 2cosx + 2cos^3x
= 4cos^3x - 3cosx
= RS
Nice!
= cos(2x + x)
= cos2xcosx - sin2xsinx
= (2cos^2 x -1)cosx - (2sinxcosx)sinx
= 2cos^3 x - cosx - 2sin^2 x (cosx)
= 2 cos^3x - cosx - 2(1 - cos^2 x)(cosx)
= 2 cos^3x - cosx - 2cosx + 2cos^3x
= 4cos^3x - 3cosx
= RS
Nice!
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