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

[phospho]

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

could anyone give me a start on how to do this?

 

[micromass]

Half angle formulas.

 

[phospho]

Half angle formulas.

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

 

[micromass]

You just got to prove that

\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{3}+1}{2\sqrt{2}}

start by squaring both sides.

 

[phospho]

You just got to prove that

\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{3}+1}{2\sqrt{2}}

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 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))

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

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, z = \sqrt{2}(cos(\frac{5\pi}{12}) + isin(\frac{5\pi}{12})) which is \sqrt{2}(m + in)^5 = ...? I could expand this using the binomial expansion but it seems unnecessary

 

[tiny-tim]

hi phospho!

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

 

[phospho]

hi phospho!


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

eh :\

 

[haruspex]

z^4 = 4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))
z = \sqrt{2}(cos(\frac{\pi}{12}+ 2k\pi) + isin(\frac{\pi}{12} + 2k\pi))

The 2kπ terms are wrong. Try again.

 

[phospho]

The 2kπ terms are wrong. Try again.

yup, silly mistake, got it thanks.