Trig Proof: Proving cos and sin of pi/12 using m and n values

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Homework Help Overview

The discussion revolves around proving the values of cosine and sine for the angle \(\frac{\pi}{12}\) in terms of specific expressions involving \(m\) and \(n\). The problem involves trigonometric identities and simplifications, particularly focusing on half-angle formulas.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of half-angle formulas and the simplification of expressions to prove the values of \(m\) and \(n\). There are attempts to relate the fourth roots of a complex number to these trigonometric values, with questions about how to express these roots in terms of \(m\) and \(n\).

Discussion Status

The discussion is active, with participants exploring various approaches to the problem. Some guidance has been provided regarding the simplification of expressions and the identification of fourth roots, though there is no explicit consensus on the methods being used.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is a focus on deriving expressions without directly providing solutions.

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prove that [itex]cos\frac{\pi}{12} = m[/itex] and [itex]sin\frac{\pi}{12} = n,[/itex] where [itex]m = \frac{\sqrt{3} + 1}{2\sqrt{2}}[/itex] and [itex]n = \frac{\sqrt{3} -1}{2\sqrt{2}}[/itex]

could anyone give me a start on how to do this?
 
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Half angle formulas.
 
micromass said:
Half angle formulas.

okay, using [itex]cos^2\frac{\pi}{12} = \dfrac{cos(\frac{\pi}{6}) + 1}{2}[/itex] I get [itex]\sqrt{\dfrac{\sqrt{3} + 2}{4}}[/itex] how could I simplify this to what they ask for (I see it's the same)
 
You just got to prove that

[tex]\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{3}+1}{2\sqrt{2}}[/tex]

start by squaring both sides.
 
micromass said:
You just got to prove that

[tex]\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{3}+1}{2\sqrt{2}}[/tex]

start by squaring both sides.

thanks, I got it -

anychance you could help with the next part?

Find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of [itex]4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))[/itex]

I started by saying
[itex]z^4 = 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))[/itex]
[itex]z = \sqrt{2}(cos(\frac{\pi}{12}+ 2k\pi) + isin(\frac{\pi}{12} + 2k\pi))[/itex]

now I get the first one easily when k = 0, but what about when k = 1, and what not, how do I get it in terms of m and n?

edit: would it be right in saying:

when k = 1, [itex]z = \sqrt{2}(cos(\frac{5\pi}{12}) + isin(\frac{5\pi}{12}))[/itex] which is [itex]\sqrt{2}(m + in)^5 = ...?[/itex] I could expand this using the binomial expansion but it seems unnecessary
 
Last edited:
hi phospho! :smile:
phospho said:
Find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of [itex]4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))[/itex]

now I get the first one easily when k = 0, but what about when k = 1, and what not …

if you have one fourth-root of a number, what are the other fourth-roots? :wink:
 
tiny-tim said:
hi phospho! :smile:


if you have one fourth-root of a number, what are the other fourth-roots? :wink:

eh :\
 
phospho said:
[itex]z^4 = 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))[/itex]
[itex]z = \sqrt{2}(cos(\frac{\pi}{12}+ 2k\pi) + isin(\frac{\pi}{12} + 2k\pi))[/itex]
The 2kπ terms are wrong. Try again.
 
haruspex said:
The 2kπ terms are wrong. Try again.

yup, silly mistake, got it thanks.
 

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