DeMoivre's Theorem Q&A: How to Do First Two Steps?

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Homework Help Overview

The discussion revolves around DeMoivre's Theorem and its application to trigonometric functions, specifically focusing on the computation of sin^6 θ and the initial steps involved in the process.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to begin the problem, mentioning an attempt to use a derived equation for sin^6 x but finding it overly complex. Other participants provide equations related to the expressions for sine and cosine in terms of complex exponentials, questioning the correctness of these equations.

Discussion Status

Some participants have offered equations that may guide the original poster in their understanding, while the original poster seeks clarification on the initial steps. There is a mix of attempts to clarify the mathematical expressions and a lack of explicit consensus on the approach to take.

Contextual Notes

The original poster indicates a struggle with the starting point of the problem, which may suggest a need for foundational understanding of the theorem's application. The discussion includes verification of mathematical expressions without resolving the overall problem.

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Homework Statement


Can somebody explain to me how the first two steps are performed?

The Attempt at a Solution


I have no idea how to start the question. I tried using an equation for sin^6 x derived by (cos x + i sinx)^6 = cos 6x+isin 6x but the solution becomes way too hard.
 

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If [itex]z = e^{i\theta}[/itex] then [tex] \cos n\theta = \frac{z^n + z^{-n}}{2} \\<br /> \sin n\theta = \frac{z^n - z^{-n}}{2i}.[/tex] Hence [tex] \sin^6 \theta = \frac{(z - z^{-1})^6}{(2i)^6} = -\frac{(z - z^{-1})^6}{64}.[/tex]
 
pasmith said:
If [itex]z = e^{i\theta}[/itex] then [tex] \cos n\theta = \frac{z^n + z^{-n}}{2} \\<br /> \sin n\theta = \frac{z^n - z^{-n}}{2i}.[/tex] Hence [tex] \sin^6 \theta = \frac{(z - z^{-1})^6}{(2i)^6} = -\frac{(z - z^{-1})^6}{64}.[/tex]
Are these equation correct?
[tex]\sin^n \theta = \frac{(z-z^{-1})^n}{(2i)^n}[/tex] and [tex]\cos^n \theta = \frac{(z+z^{-1})^n}{2^n}[/tex]
 
Last edited:
Yes.
 

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