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Why is sin(3x) = 3sin(x)cos(x)cos(x) - sin(x)sin(x)sin(x)
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Expand Sin(3x) = Sin(2x+x) and in the answer expand the Sin(2x) =Sin (x+x) and Cos(2x) = Cos(x+x) and you will see...
Assuming you know \[ sin(a+b) = sin(a)cos(b)+cos(a)sin(b) \] and \[ cos(a+b) = cos(a)cos(b)-sin(a)sin(b) \] We get \[ sin(3x) = sin(2x+x) = sin(2x)cos(x)+cos(2x)sin(x) = \] \[ = sin(x+x)cos(x)+cos(x+x)sin(x) = \] \[ = [2sin(x)cos(x)]cos(x)+[cos(x)cos(x)-sin(x)sin(x)]sin(x) \] \[ = 3sin(x)cos(x)cos(x)-sin(x)sin(x)sin(x) = 3\sin(x)\cos^2(x)-\sin^3(x) \]
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