Linearizing $\sin^4\theta$

Expansion.
$$\sin^4\theta=\frac{1}{16}(e^{i\theta}-e^{-i\theta})^4=\frac{1}{16}(e^{4i\theta}-4e^{2i\theta}+6-4e^{-2i\theta}+e^{-4i\theta}).$$

Regroup.
$$=\frac{1}{16}[(e^{4i\theta}+e^{-4i\theta})-4(e^{2i\theta}+e^{-2i\theta})+6]=\frac{1}{16}[2\cos4\theta-8\cos2\theta+6]$$
$$=\frac{\cos4\theta-4\cos2\theta+3}{8}.$$

Conclusion. $\sin^4\theta=\frac{\cos4\theta-4\cos2\theta+3}{8}$.