Let $$\omega = \cos\left({10^o} \right) + i \sin\left(10^o\right)$$ Then $$\omega^3 = \cos(30^o) + i \sin(10^o) = \frac{\sqrt{3}+i}{2}$$ and $$ \omega = \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} $$ You have to be careful with branches of the cube root, but this turns out to be correct if you use the principal branch. Then $\sin(10^o)$ is the imaginary part of this, i.e. $$ \sin(10^o) = \frac{\omega - \overline{\omega}}{2i} = -\frac{i}{2} \left( \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} - \left(\frac{\sqrt{3}-i}{2}\right)^{1/3}\right)$$

Let $$\omega = \cos\left({10^o} \right) + i \sin\left(10^o\right)$$ Then $$\omega^3 = \cos(30^o) + i \sin(30^o) = \frac{\sqrt{3}+i}{2}$$ and $$ \omega = \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} $$ You have to be careful with branches of the cube root, but this turns out to be correct if you use the principal branch. Then $\sin(10^o)$ is the imaginary part of this, i.e. $$ \sin(10^o) = \frac{\omega - \overline{\omega}}{2i} = -\frac{i}{2} \left( \left(\frac{\sqrt{3}+i}{2}\right)^{1/3} - \left(\frac{\sqrt{3}-i}{2}\right)^{1/3}\right)$$