double-angle identity proof???

Question

how do i prove this
2 cos^2 A - 1 = cos(2A)


using complex numbers where
cos A = ( e^-iA + e^iA ) / 2

and

sin A = ( e^iA - e^-iA ) / 2i


plz help

Answer 1

cos(A) = ( e^-iA + e^iA ) / 2

cos^2(A) = ( e^-iA + e^iA )^2 / 4
= [e^(-2iA) + 2e^(iA)e^(-iA) + e^(2iA)] / 4
= [e^(-2iA) + 2e^(iA-iA) + e^(2iA)] / 4
= [e^(-2iA) + 2e^0 + e^(2iA)] / 4
= [e^(-2iA) + 2 + e^(2iA)] / 4
and so

2cos^2(A) - 1 = 2*[e^(-2iA) + 2 + e^(2iA)] / 4 - 1
= [e^(-2iA) + 2 + e^(2iA)] / 2 - 2/2
= [e^(-2iA) + 2 + e^(2iA) - 2] / 2
= [e^(-2iA) + e^(2iA)] / 2
= cos(2A)

Answer 2

By manual computation

2 cos² A - 1
2 ((e^(-iA) + e^(iA))/2)² - 1
2 (e^(-iA))² + 2e^(-iA)e^(iA) + (e^(iA))²)/4 - 1
(e^(-2iA) + 2 + e^(2iA))/2 - 1
(e^(-2iA) + e^(2iA))/2 + 2/2 - 1
(e^(-2iA) + e^(2iA))/2
cos (2A)

Q.E.D.