I think you mean in terms of [tex]\sin(x)[/tex]?
Recall Euler's identity
[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]
and de Moivre's theorem
[tex]\left(e^{ix}\right)^n = \left(\cos(x) + i \sin(x)\right)^n = \cos(nx) + i \sin(nx) = e^{inx}[/tex]
where [tex]i=\sqrt{-1}[/tex].
It follows that
[tex]\sin(4x) = \mathrm{Im}\left(\cos(x) + i \sin(x)\right)^4[/tex]
By the binomial theorem, expanding the right side gives
[tex]\cos^4(x) + 4i \cos^3(x) \sin(x) - 6\cos^2(x) \sin^2(x) - 4i \cos(x) \sin^3(x) + \sin^4(x)[/tex]
and so
[tex]\sin(4x) = 4\cos^3(x) \sin(x) - 4 \cos(x) \sin^3(x)[/tex]
We can factorize this as
[tex]\sin(4x) = 4 \cos(x) \sin(x) \left(\cos^2(x) - \sin^2(x)\right)[/tex]
and using the Pythagorean identity
[tex]\cos^2(x)+\sin^2(x) = 1 \implies \cos(x) = \pm \sqrt{1-\sin^2(x)}[/tex]
this reduces to
[tex]\sin(4x) = \pm 4 \sqrt{1-\sin^2(x)} \sin(x) (1 - 2 \sin^2(x))[/tex]
