Trigonometric double angle identities?

Question

Create a formula from sin(3x) in terms of sin(x).

Answer 1

Before we get into that, our identities we're working with is
sin(a + b) = sin(a)cos(b) + sin(b)cos(a), and
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2sin^2(x)

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

Expanding everything out,

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

Using the identity cos^2(x) = 1 - sin^2(x),

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