How to prove #sin(4x) = 8 (cos^3)x sin x − 4 cos x sin x#? With sin(4x) = Im(e^(i4x))

1 Answer
Dec 14, 2017

See the proof below

Explanation:

We apply Demoivre's Theorem

#(cosx+isinx)^n=cos(nx)+isin(nx)#

Let #n=4#

#cos(4x)+isin(4x)=e^(i4x)=(cosx+isinx)^4#

Therefore,

#sin(4x)=Im(e^(i4x))=Im(cosx+isinx)^4#

Develop #(cosx+isinx)^4# by the Binomial theorem

#(cosx+isinx)^4=cos^4x+4cos^3x(isinx)+6cos^2x(isinx)^2+4cosx(isinx)^3+(isinx)^4#

#=cos^4x-6cos^2xsin^2x+sin^2x+i(4cos^3xsinx-4cosxsin^3x)#

Therefore,

#Im(cosx+isinx)^4=4cos^3xsinx-4cosxsin^3x#

#=4cos^3xsinx-4cosxsinx(1-cos^2x)#

#=4cos^3xsinx-4cosxsinx+4cos^3xsinx#

#=8cos^3xsinx-4cosxsinx#

#QED#