How do you prove #sin3x = 3sinx - 4sin^3x#?

2 Answers

Since the angle sum formula of sine is:

#sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta#,

and the double angle formula of cosine:

#cos(2alpha)=cos^2alpha-sin^2alpha=2cos^2alpha-1=1-2sin^2alpha#

then:

#sin(3x)=sin(2x+x)=sin(2x)cosx+cos(2x)sinx=#

#=(2sinxcosx)*cosx+(1-2sin^2x)sinx=#

#=2sinxcos^2x+sinx-2sin^3x=#

#=2sinx(1-sin^2x)+sinx-2sin^3x=#

#=2sinx-2sin^3x+sinx-2sin^3x=#

#=3sinx-4sin^3x#.

Mar 23, 2018

Answer:

we can use DeMoivre's theorem

Explanation:

#(costheta+isintheta)^n=cosntheta+isinntheta--(1)#

since we want #sin^3x# we will expand #(cosx+isinx)^3#

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

#(cosx+isinx)^3=cos^3x-3cosxsin^2x+i(3cos^2xsinx-sin^3x)#

from#(1)#

#cos3x+isin3x=cos^3x-3cosxsin^2x+i(3cos^2xsinx-sin^3x)#

equating imaginary parts

#sin3x=3cos^2xsinx-sin^3x#

but #cos^2x=1-sin^2x#

#sin3x=3(1-sin^2x)sinx-sin^3x#

#sin3x=3sinx-3sin^2xsinx-sin^3x#

#sin3x=3sinx-4sin^3x#

as required