Find an expression in terms of powers of cos(x) for cos(5x)

De Moivre's theorem states that e^ix= cos(x) + isin(x) or that e^i5x= cos(5x)+ isin(5x). If the real components of both sides of this equation is taken we can see that : cos(5x) = Re[e^i5x ] where Re means take the real component of
Also e^i5x= e^ix*5 =(cos(x) + isin(x))⁵ using laws of index multiplication.
Therefore cos(5x) = Re[(cos(x) + isin(x))⁵ ]For easy of writing let us use notation c= cos(x) and s= sin(x). We can thus write cos(5x) =Re[(c+is)⁵]
Expanding the bracket using binomial theorem cos(5x) = c⁵-10c³s²+5cs⁴
Pythagoras's identity states sin²x + cos²x =1Rearranging we can write s²=1-c²
Substituting this expression for s² we get cos(5x) = c⁵-10c³(1-c²)+5c(1-c²)²Expanding the brackets and gathering like powers of cos xwe get cos(5x)= 16c⁵-20c³+5cChanging back notation we can writecos(5x)= 16cos⁵(x)-20cos³(x)+5cos(x)