MQ15---Trigonometry

Since
(1 + sin x + i cos x) / (1 + sin x - i cos x)
= [1 + cos (π/2 - x) + i sin (π/2 - x)] / [1 + cos (π/2 - x) - i sin (π/2 - x)]
= [2 cos ^2 (π/4 - x/2) + 2i sin (π/4 - x/2) cos (π/4 - x/2)] / [2 cos ^2 (π/4 - x/2) - 2i sin (π/4 - x/2) cos (π/4 - x/2)]
= [cos (π/4 - x/2) + i sin (π/4 - x/2)] / [cos (π/4 - x/2) - i sin (π/4 - x/2)]
= cos (π/2 - x) + i sin (π/2 - x)
Therefore,
[(1 + sin x + i cos x) / (1 + sin x - i cos x)]^n
= [cos (π/2 - x) + i sin (π/2 - x)]^n
= cos (nπ/2 - nx) + i sin (nπ/2 - nx)