What property was used in this sine transformation? I have this expression:
$$
ψ(χ) = A\sin^3(\frac{πχ}{α})
$$
And somehow the book i read equalizes the previous equation to this one:
$$
ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})]
$$
What trigonometric identity was used to make this possible?

The identity is 
$$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$
 Proof:
\begin{align*} 
 4\sin^3x-3\sin x &=  \sin x(4\sin^2x-3)
 \\ 
 &=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity} 
\\ 
 &=\sin x(2\cos^2x-\cos(2x)-4\cos^2x) \quad \text{Double angle cosine} 
\\ 
 &=-\sin x(2\cos^2x+\cos(2x))
 \\ 
 &=-[(2\sin x\cos x)\cos x+\sin x\cos(2x)]
\\ 
 &=-[\sin (2x)\cos x+\sin x\cos(2x)]\quad \text{Double angle sine}
\\ 
 &=-\sin (2x+x)\quad \text{identity sum sine}
\\ 
 &=-\sin(3x)
\end{align*}
