Need to prove the Identity sin(3x)=sinx(4cos²x-1)?

Question

Ok can you go through that step by step because I feel like your skiping some steps

Answer 1

dealing with the left hand side...
you know the sum properties, so sin(3x) can be divided into sin(x+2x).

sin(x+2x)
=sinxcos2x + sin2xcosx

you also know that sin2x= 2sinxcosx
so...

=sinxcos2x + (2 sinxcosx)cosx

now cos2x can be changed to (cos^2)x-(sin^2)x

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

now you can change (sin^2)x to 1-(cos^2)x using the identities. and simplifying gives you this equation:

=sinx(2(cos^2)x-1) + (2sinxcosx)cosx

now factor out the sin x

=sinx[(2(cos^2)x-1)+(2cosxcosx)]

the two cosx in the 2cosxcosx combine:

=sinx[2(cos^2)x-1+2(cos^2)x]

now just simplify by adding the two cos^2

=sinx[4(cos^2)-1]

And there's your answer! Just as an added tip... its usually easier to prove identities by working with the simpler side and trying to derive the more complicated side

Answer 2

from LHS

sin 3x=sin(2x+x)
=sin2xcosx+cos2xsinx
=2sinxcos^2 x - (2cos^2 x -1)sinx
=2cos^2 x sinx + 2cos^2 x sinx - sin x
=4sin x cos^2 x -sin x
=sin x(4cos^2 x -1)

Answer 3

Just use the fact that sin(a+b) = sin a cos b + sin b cos a.
sin(3x)
= sin(2x + x)
= sin 2x cos x + sin x cos 2x
= (2 sin x cos x) cos x + sin x (2(cos x)^2 - 1)
= 2 sin x (cos x)^2 + 2 sin x (cos x)^2 - sin x
= sin x(4 (cos x)^2 - 1).

Answer 4

sin(3x) = (sin(x))(4cos(x)^2 - 1)

sin(3x) = sin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)
sin(3x) = 2sin(x)cos(x)cos(x) + sin(x)(2cos(x)^2 - 1)
sin(3x) = 2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x)
sin(3x) = 4sin(x)cos(x)^2 - sin(x)
sin(3x) = (sin(x))(4cos(x)^2 - 1)

Therefore

sin(3x) = (sin(x))(4cos(x)^2 - 1)

for more info, go to http://www.math.com/tables/trig/identities.htm