Show that sin 3x = 3 sin x - 4sin^3 x ?

# sin (A+-B) = sin A cos B +- cos A sin B #
# sin 2A = 2sin A cos A #
# cos 2A = cos^2 A - sin^2 A #
# sin^2 A + cos^2 A = 1 #

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

If you have encountered Euler's Formula then it works out like this:

Euler's Formula gives us: #cos x + i sin x = e^(ix)#

So:

#cos(3x)+i sin(3x) = e^(i3x) = (e^(ix))^3 = (cos(x)+i sin(x))^3#

#=cos^3(x)+3cos^2(x) i sin(x) + 3 cos(x) i^2 sin^2(x) + i^3 sin^3(x)#

#=(cos^3(x)-3 cos(x)sin^2(x)) + i(3 cos^2(x) sin(x)-sin^3(x))#

Equating the imaginary parts of both ends, we get:

#sin(3x) = 3 cos^2(x)sin(x)-sin^3(x)#

#= sin(x)(3 cos^2(x)-sin^2(x))#

#=sin(x)(3(1 - sin^2(x))-sin^2(x))# #color(green)([[ "using: " sin^2(x)+cos^2(x) = 1 ]])#

#=sin(x)(3-4sin^2(x))#

#=3sin(x)-4sin^3(x)#

Equating the real parts of both ends, we also get:

#cos(3x) = cos^3(x)-3 cos(x)sin^2(x)#

#=cos(x)(cos^2(x)-3sin^2(x))#

#=cos(x)(cos^2(x)-3(1-cos^2(x)))#

#=cos(x)(4 cos^2(x)-3)#

#=4cos^3(x)-3cos(x)#