Let $z=\cos\theta+i\sin\theta$. Then by De Moivre's Law, we have $$z^3 = (\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$$ Meanwhile, binomial expansion yields \begin{align} z^3 &= (\cos\theta + i\sin\theta)^3 \\ &= \cos^3\theta + 3\cdot\cos^2\theta\cdot(i\sin\theta) + 3\cdot\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3\\ &= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(\cos^2\theta\sin\theta - \sin^3\theta)\\ &= (\cos^3\theta - 3\cos\theta(1-\cos^2\theta)+i((1-\sin^2\theta)\sin\theta-\sin^3\theta)\\ &= (4\cos^3\theta - 3\cos\theta) + i (3\sin\theta - 4\sin^3\theta) \end{align} These two have to be the same. Setting both real parts and imaginary parts equal leads to the desired result.