If I understand what you're asking...:
With the first one, you would use the fact that
z+z1=2cosx. Whatever your value of n is, you would raise both sides of the equation to that power, and binomially expand the first part, collect the like terms and simplify. That way, your cos^n(x) will now be expressed in terms of cos(nx).
With the latter, you would use de Moivre's Theorem to rewrite cos(nx) as [cos(x) + isin(x)]^n, and binomially expand that. Then, you'd equate the real terms, if you wanted it in terms of cos(x), or you'd equate the imaginary terms if you wanted it in terms of sin(x).
(If you're asking in terms of Taylor series, then you'd use the standard result for cos(x) and replace all of the x's with (nx), when expanding the second one. And you would expand the first using Taylor's series by writing cos^n(x) as [cos(x)]^n and calculating the derivatives of that up to what you want to expand to and substituting it in.)
Hope it helps and I haven't confused you too much, lol.