Propiedad:
Recordemos que $\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right) = \frac{1+\tan(\theta/2)}{1-\tan(\theta/2)} = \sec \theta + \tan \theta = \frac{1+\sin \theta}{\cos \theta}$.
También se cumple que $\tan^2\left(\frac{\theta}{2} + \frac{\pi}{4}\right) = \frac{1+\sin \theta}{1-\sin \theta}$.

Desarrollo:
Sea $T_\alpha = \tan\left(\frac{\alpha}{2} + \frac{\pi}{4}\right)$ y $T_\beta = \tan\left(\frac{\beta}{2} + \frac{\pi}{4}\right)$. El enunciado dice $T_\alpha^3 = T_\beta$.
Sabemos que $\sin \beta = \frac{T_\beta^2 - 1}{T_\beta^2 + 1}$.
Sustituyendo $T_\beta = T_\alpha^3$:
$$ \sin \beta = \frac{T_\alpha^6 - 1}{T_\alpha^6 + 1} $$
Sea $x = \sin \alpha$, entonces $T_\alpha^2 = \frac{1+x}{1-x}$.
Sustituimos $T_\alpha^6 = \left(\frac{1+x}{1-x}\right)^3$:
$$ \sin \beta = \frac{\frac{(1+x)^3}{(1-x)^3} - 1}{\frac{(1+x)^3}{(1-x)^3} + 1} = \frac{(1+x)^3 - (1-x)^3}{(1+x)^3 + (1-x)^3} $$
Desarrollando los cubos:
$$ (1+x)^3 = 1 + 3x + 3x^2 + x^3 $$
$$ (1-x)^3 = 1 - 3x + 3x^2 - x^3 $$
Numerador: $(1 + 3x + 3x^2 + x^3) - (1 - 3x + 3x^2 - x^3) = 6x + 2x^3 = 2x(3 + x^2)$.
Denominador: $(1 + 3x + 3x^2 + x^3) + (1 - 3x + 3x^2 - x^3) = 2 + 6x^2 = 2(1 + 3x^2)$.
$$ \sin \beta = \frac{2x(3 + x^2)}{2(1 + 3x^2)} = \frac{x(3 + x^2)}{1 + 3x^2} $$
Sustituyendo $x = \sin \alpha$:

Resultado:
$$ \boxed{\sin \beta = \frac{(3 + \sin^2 \alpha) \sin \alpha}{1 + 3 \sin^2 \alpha}} $$