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

2 Answers
Massimiliano · mason m · Jacobi J.
Apr 26, 2015

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.

sjc · Ratnaker Mehta
Mar 23, 2018

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

Related questions
See all questions in Proving Identities
Impact of this question
127439 views around the world
You can reuse this answer
Creative Commons License