Question

[Pre-Calc]

Answer 1

sin(α+β)=sinαcosβ+cosαsinβ,

and the double angle formula of cosine:

cos(2α)=cos2α−sin2α=2cos2α−1=1−2sin2α

then:

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

=(2sinxcosx)⋅cosx+(1−2sin2x)sinx=

=2sinxcos2x+sinx−2sin3x=

=2sinx(1−sin2x)+sinx−2sin3x=

=2sinx−2sin3x+sinx−2sin3x=

=3sinx−4sin3x.

cos3x
=cos(x+2x)
=cosxcos2x−sinxsin2x using cos(A+B)=cosAcosB-sinAsinB
=cosx(2cos^2x−1)−sinx(2sinxcosx) using cos2A=2cos^2A-1 and sin2A=2sinAcosA
=2cos^3x−cosx−2sin^2xcosx 
=2cos^3x−cosx−2(1−cos^2x)cosx  using sin^2x+cos^2x=1
=2cos^3x−cosx−2(cosx−cos3^x) [open the brackets]
=2cos^3x−cosx−2cosx+2cos^3x
=4cos^3x−3cosx [collect terms]