Given z=cosx+isinx, show cosx=1/2(z+1/z)

We first consider 1/z=(cosx+isinx)^(-1). Application of De Moivre's theorem for integer n: (cosx+isinx)^(n)=cosnx+isinnx yields the result 1/z=cosx-isinx. Addition of the two forms z and z^(-1) steers us to the result, albeit with this being double the result.