Proving cos(π/12) and sin(π/12) equal specific expressions

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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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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.