# Trig proof

1. Nov 16, 2012

### 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?

2. Nov 16, 2012

### micromass

Staff Emeritus
Half angle formulas.

3. Nov 16, 2012

### phospho

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)

4. Nov 16, 2012

### micromass

Staff Emeritus
You just got to prove that

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

start by squaring both sides.

5. Nov 16, 2012

### phospho

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

Last edited: Nov 16, 2012
6. Nov 16, 2012

### tiny-tim

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

7. Nov 16, 2012

### phospho

eh :\

8. Nov 17, 2012

### haruspex

The 2kπ terms are wrong. Try again.

9. Nov 18, 2012

### phospho

yup, silly mistake, got it thanks.