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Problem If θ ∈ R \theta \in \mathbb{R} θ ∈ R and i = − 1 i = \sqrt{-1} i = − 1 , then ( 1 + sin θ + i cos θ 1 + sin θ − i cos θ ) n \left(\frac{1 + \sin \theta + i \cos \theta}{1 + \sin \theta - i \cos \theta}\right)^n ( 1 + s i n θ − i c o s θ 1 + s i n θ + i c o s θ ) n is equal to
a. cos ( n π 2 − n θ ) + i sin ( n π 2 − n θ ) \cos\left(\frac{n\pi}{2} - n\theta\right) + i \sin\left(\frac{n\pi}{2} - n\theta\right) cos ( 2 nπ − n θ ) + i sin ( 2 nπ − n θ )
b. cos ( n π 2 + n θ ) + i sin ( n π 2 + n θ ) \cos\left(\frac{n\pi}{2} + n\theta\right) + i \sin\left(\frac{n\pi}{2} + n\theta\right) cos ( 2 nπ + n θ ) + i sin ( 2 nπ + n θ )
c. sin ( n π 2 − n θ ) + i cos ( n π 2 − n θ ) \sin\left(\frac{n\pi}{2} - n\theta\right) + i \cos\left(\frac{n\pi}{2} - n\theta\right) sin ( 2 nπ − n θ ) + i cos ( 2 nπ − n θ )
d. cos ( n ( π 2 + 2 θ ) ) + i sin ( n ( π 2 + 2 θ ) ) \cos\left(n\left(\frac{\pi}{2} + 2\theta\right)\right) + i \sin\left(n\left(\frac{\pi}{2} + 2\theta\right)\right) cos ( n ( 2 π + 2 θ ) ) + i sin ( n ( 2 π + 2 θ ) )
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