# Answer to Question #16168 in Trigonometry for soni sharma

Question #16168

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

[sin(α+nθ)-e^iαsin nθ]=e^ -inθ sinα

[sin(α+nθ)-e^iαsin nθ]=e^ -inθ sinα

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