Prove $ [\cos(\theta + \frac{\pi}{3}) + i\sin(\theta + \frac{\pi}{3})] ^6 = \cos(6\theta) +i\sin(6\theta)$

Question

I am really unsure how to prove the above. I understand any the De Moivre theorem applies to any $\theta$ but I don't know in this instance how to simply and prove it.

Answer 1

$6(\theta+\frac{\pi}{3})=6\theta+2\pi$

What's $\cos(\alpha+2\pi)$? What about the sine?

Final hint: De Moivre.

Answer 2

$(\exp i(\theta+\pi/3))^6=\exp i(6\theta +2\pi)=\exp 6i\theta$ because $\exp 2\pi i=1$.