Solution
= Solution
Take the real part of <De Moivre's theorem>. Put $c=\cos\theta$ and use $\sin^2\theta=1-c^2$:
$$
\cos5\theta=c^5-10c^3(1-c^2)+5c(1-c^2)^2=\boxed{16\cos^5\theta-20\cos^3\theta+5\cos\theta}.
$$
= Solution
Take the real part of <De Moivre's theorem>. Put $c=\cos\theta$ and use $\sin^2\theta=1-c^2$:
$$
\cos5\theta=c^5-10c^3(1-c^2)+5c(1-c^2)^2=\boxed{16\cos^5\theta-20\cos^3\theta+5\cos\theta}.
$$