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
1
Expert's answer
2012-10-15T12:02:03-0400
The answer to the question is available in the PDF file https://www.assignmentexpert.com/https://www.assignmentexpert.com/homework-answers/mathematics-answer-16168.pdf

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