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`(cos theta+i sin theta)^n = cos ntheta +i sin ntheta`
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If z = cos θ + i sin θ then z^n+1/z^n = 2 cos nθ and z^n -1/z^n = 2 i sin nθ || Complex Numbers
`(cos theta+i sin theta)^n = cos ntheta +i sin ntheta`
The value of 〖(sin〖θ+i cosθ 〗)〗^n is
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