Question

de moivre's theorem if a=cos 2alpha + i sin 2alpha b= cos 2beta + i sin 2beta prove, a-b/a+b = i tan (alpha-beta)

Answer 1

i think we use just algebraic games to solve this problem
a-b = cos(2A) + i sin(2A) - cos(2B) - i sin(2B)
a-b = [cos(2A) - cos(2B)] + i[sin(2A) - sin(2B)]
a-b = [-2sin(A+B)sin(A-B)] + i[2cos(A+B)sin(A-B)]
a-b = [(-2)sin(A-B)][sin(A+B) - i cos(A+B)]
i(a-b) = [(-2)sin(A-B)][i sin(A+B) + cos(A+B)]
a-b = [(-2/i)sin(A-B)][cos(A+B) + i sin(A+B)]
try it for a+b