Derive the identity for sin 3x in terms of sin x?

Answer 1

sin(2x)=2sinxcosx
cos(2x)=cos^2(x)-sin^2(x)

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

Answer 2

sin 3x = sin (x+2x)
= sin x cos 2x + sin 2x cos x
= sin x (1 - 2 (sin x)^2) + (2 sin x cos x) cos x
= sin x (1 - 2(sin x)^2) + 2 sin x (cos x)^2
= sin x (1 - 2(sin x)^2) + 2 sin x (1 - (sin x)^2)
You can expand to get
= sin x - 2(sin x)^3 + 2 sin x - 2(sin x)^3
= 3 sin x - 4 (sin x)^3.

Answer 3

Sin 3x
=Sin (2x+x)
=sin 2x cos x +cos 2x sin x
=sin (x+x) cos x +cos (x+x) sin x
=(sin x cos x + sin x cos x)cos x + (cos x cos x - sin x sin x) sin x
=2*sin x (cosx)^2+ (cos x)^2 sin x - (sin x)^3
=3 * sin x *[ 1 - (sin x)^2] - (sin x)^3
= 3 sin x - 4(sin x)^3