Help Evaluating Complex Numbers

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Homework Help Overview

The discussion revolves around evaluating the product of sine functions, specifically sin(π/7)sin(2π/7)sin(3π/7), using complex numbers and roots of unity. Participants are exploring connections between sine and cosine products derived from the equation z^7 - 1.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between sine and cosine products and question how to derive the sine product from the cosine product already evaluated. There is mention of using complex exponentials and roots of unity to approach the problem.

Discussion Status

Some participants have confirmed the correctness of the cosine product evaluation, while others are questioning how to adapt that method for the sine product. Suggestions for alternative approaches, such as expressing sine in terms of complex exponentials, have been introduced.

Contextual Notes

There is an emphasis on the need to connect the sine product evaluation to the established cosine product result, with participants noting the potential challenges in doing so. The original poster's attempt at a solution is based on a specific equation, which may limit the applicability of their findings to the sine case.

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Homework Statement



Evaluate:

Homework Equations



[tex]sin\frac{\pi }{7}.sin\frac{2\pi }{7}.sin\frac{3\pi }{7}[/tex]

The Attempt at a Solution



[tex]Using[/tex] [tex]z^{7}-1[/tex] [tex]got:[/tex]
[tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
 
Last edited:
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Seems correct!
 
micromass said:
Seems correct!

But my solution is for the cosine product. How can I find the solution to the question from what I got?
 
Can't you repeat much of the same proof? Where does it fail?
 
The way I got: [tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex] is by solving the equation [tex]z^{7}-1=0[/tex]
[tex]z=1,cis\pm\frac{2\pi }{7},cis\pm\frac{4\pi }{7},cis\pm\frac{6\pi }{7}[/tex]
[tex]w+w^{-1}=cis\frac{2\pi }{7}+cis\frac{-2\pi }{7}=2cos\frac{2\pi }{7}[/tex]
Similarly;
[tex]w^{2}+w^{-2}=-2cos\frac{3\pi }{7}[/tex]
[tex]w^{3}+w^{-3}=-2cos\frac{\pi }{7}[/tex]
[tex](w+w^{-1})(w^{2}+w^{-2})(w^{3}+w^{-3})=[/tex]
[tex]w^{6}+1+w^{2}+w^{3}+w^{4}+w^{5}+1+w=1[/tex]
, since [tex]w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1=0 \therefore cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
So I don't think I can do the same to get the produce for sine.
 
It might be easier to determine

[tex]\prod_{k=0}^6 \sin(k\pi/7)[/tex]

Do this by expressing the sine as complex exponentials. Also try to make use of the following formula

[tex]\frac{z^7-1}{z-1}=\prod_{k=1}^6 (1- \zeta^k)[/tex]

with [itex]\zeta = \cos(\pi/7) + i\sin(\pi/7)[/itex]. Let z go to 1.
 

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