Solution
= Solution
By <De Moivre's theorem>, $(\cos\theta+i\sin\theta)^4=\cos4\theta+i\sin4\theta$. Take real parts of the binomial expansion:
$$
\cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta.
$$
Eliminate $\sin^2\theta$ using $1-\cos^2\theta$ and collect powers:
$$
\boxed{\cos4\theta=8\cos^4\theta-8\cos^2\theta+1.}
$$