3.390. Derive the formula $\cos (n+1) \alpha=2 \cos \alpha \cos n \alpha-\cos (n-1) \alpha$, where $n$ is any real number, and use it to express $\cos 3 \alpha$ and $\cos 4 \alpha$ as polynomials in $\cos \alpha$.
Solution.

$\cos (n+1) \alpha=\cos (n \alpha+\alpha)=[\cos (x+y)=\cos x \cos y-\sin x \sin y]=$

$=\cos n \alpha \cos \alpha-\sin n \alpha \sin \alpha=2 \cos n \alpha \cos \alpha-\cos n \alpha \cos \alpha-\sin n \alpha \sin \alpha:=$

$=2 \cos n \alpha \cos \alpha-(\cos n \alpha \cos \alpha+\sin n \alpha \sin \alpha)=$

$=2 \cos n \alpha \cos \alpha-\cos (n \alpha-\alpha)=2 \cos n \alpha \cos \alpha-\cos (n-1) \alpha$;

$\cos 3 \alpha=\cos (2+1) \alpha=2 \cos \alpha \cos 2 \alpha-\cos \alpha=$

$=2 \cos \alpha\left(2 \cos ^{2} \alpha-1\right)-\cos \alpha=4 \cos ^{3} \alpha-2 \cos \alpha-\cos \alpha=$

$=4 \cos ^{3} \alpha-3 \cos \alpha ;$

$\cos 4 \alpha=\cos (3+1) \alpha=2 \cos \alpha \cos 3 \alpha-\cos 2 \alpha=$

$=2 \cos \alpha\left(4 \cos ^{3} \alpha-3 \cos \alpha\right)-\left(2 \cos ^{2} \alpha-1\right)=$

$=8 \cos ^{4} \alpha-6 \cos ^{2} \alpha-2 \cos ^{2} \alpha+1==8 \cos ^{4} \alpha-8 \cos ^{2} \alpha+1$.

Answer: $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha ; \cos 4 \alpha=8 \cos ^{4} \alpha-8 \cos ^{2} \alpha+1$.