Use de Moivre's theorem to show that when n is a positive integer then
$cosnx$ = $\cos^nx$- ${n \choose 2}\cos^{n-2}x$$\sin^2x$ +${n \choose 4}\cos^{n-4}x\sin^4x$ +...... $(-1)^{n/2} \sin^nx$
Deduce that 1 - 3${n \choose 2}$ +$3^2{n\choose 4}$ -..... +$(-3)^{n/2}$=$2^ncos(n\pi /3)$
Can you all show me how to get the last term for $cosnx$??
Simply put x=$\frac{\pi}{3}$ it will change cosx=$\frac{1}{2}$ and sinx=$\frac{\sqrt3}{2}$. Take $2^n$ on RHS