De Moivre's Theorem Simplification

Question

Prove that $$(1 + \operatorname{cis}(x))^k + (1 + \operatorname{cis}(-x))^k = 2^{k+1}\cos\left(\frac{kx}{2}\right)\cos^k\left(\frac{x}{2}\right).$$

Any guidance would be greatly appreciated.

Thanks.

Answer 1

Start by dividing both sides by $2^k$
Then calculate $\frac{(1+cis(x))^k}{2^k} = cis(\frac{x}{2})^k\frac{(cis(\frac{-x}{2})+cis(\frac{x}{2}))^k}{2^k}=cis(\frac{x}{2})^kcos^k(\frac{x}{2})$
The same for $\frac{(1+cis(-x))^k}{2^k} = cis(\frac{-x}{2})^kcos^k(\frac{x}{2})$
Sum both terms and factor by $cos^k(\frac{x}{2})$
That gives you $cos^k(\frac{x}{2})(cis(\frac{x}{2})^k + cis(\frac{-x}{2})^k) = cos^k(\frac{x}{2})(cis(\frac{kx}{2}) + cis(\frac{-kx}{2}))$ using Moivre formula beacuse $k\in\Bbb N$
Finally $cis(\frac{kx}{2}) + cis(\frac{-kx}{2}) = 2cos(\frac{kx}{2})$
Thus $$\frac{(1+cis(x))^k+(1+cis(-x))^k}{2^k} = 2cos(\frac{kx}{2})cos^k(\frac{x}{2})$$

Hope it helps