Question

prove that - [sin(α+θ)-e^iαsinθ]^n =e^ -inθ sinα^n
[sin(α+nθ)-e^iαsin nθ]=e^ -inθ sinα

Answer 1

1) $\left(\sin (u + v) - e^{iu}\sin v\right)^n = \left(\sin u\cos v + \cos u\sin v - \cos u\sin v - i\sin u\sin v\right)^n =$

$$= \sin^ {n} u (\cos v - i \sin v) ^ {n} = e ^ {- i n v} \sin^ {n} u$$

2) $\sin (u + nv) - e^{iu}\sin nv = \frac{1}{2i}\Big(e^{i(u + nv)} - e^{-i(u + nv)} - e^{i(u + nv)} + e^{i(u - nv)}\Big) = \frac{e^{-inv}}{2i}\Big(e^{iu} - e^{-iu}\Big) = e^{-inv}\sin u$