Trig3.8

Prove that $16\cos^5 \theta-20\cos^3 \theta+5\cos \theta=\cos 5\theta\,$

RHS= $\cos 5\theta=\cos (4\theta+\theta)=\cos 4\theta \cos \theta-\sin 4\theta \sin \theta\,$

$(2\cos^2 2\theta -1)\cos \theta-2\sin 2\theta \cos 2\theta \sin \theta\,$

$2(2\cos^2 \theta-1)^2 \cos \theta-\cos \theta-4\sin \theta \cos \theta (2\cos^2 \theta-1)\sin \theta\,$

$2\cos \theta(4\cos^4 \theta+1-4\cos^2 \theta)-\cos \theta-4\cos \theta(1-\cos^2 \theta)(2\cos^2 \theta-1)\,$

$8\cos^5 \theta-8\cos^3 \theta+2\cos \theta-\cos \theta-4\cos \theta(3\cos^2 \theta-2\cos^4 \theta-1)\,$

$8\cos^5 \theta-8\cos^3 \theta+\cos \theta-12\cos^3 \theta+8\cos^5 \theta+4\cos \theta\,$

$16\cos^5 \theta-20\cos^3 \theta+5\cos \theta\,$ =LHS