# Finding cos in radians without calculator? Help?

1. Nov 5, 2012

### mathFun

Hey guys this question is gonna sound pretty dumb, but I never learned how to do this. What if I want to find say, cos(pi/5) but I don't have a calculator? Even if I did have a calculator, I want it in nice values, like square roots, rather than decimals anyway.

If I have something like cos(pi/8) I think I know how to find something like that using double angle formula cos(2x)=2cos2x-1, because I know the value of cos(4pi) and cos(2pi), but when I have an odd value I'm stuck at what to do.

Is there any simple algorithmic process I can use? Someone mentioned to me about finding complex roots, but I'm not sure how that works?

2. Nov 5, 2012

### micromass

Staff Emeritus
This wiki article might be useful: http://en.wikipedia.org/wiki/Exact_trigonometric_constants
It shows how to find things like $\cos(\pi/5)$ using geometry.

Note that it is not always possible to find explicit value of cosines. For exampe, the innocent looking $\cos(\pi/9)$ cannot be explicitely found using elementary operations and roots (unless I'm being stupid right now).

3. Nov 5, 2012

### arildno

Well, pi/5 lies between pi/4 and pi/6, doesn't it?
The midpoint between these to latter values is 5pi/24 which is slightly greater than pi/5

calculating the relevant values of sine and cosine for 5pi/24, you may find the midpoint between 5pi/24 and pi/6, which is 9pi/48, somewhat less than pi/5
(And then, the midpoint between 5pi/24 and 9pi/48 is 19pi/96, very close to, but slightly less than pi/5)

And so on.

Utilizing this bisection method and the half-angle formula, you can eke yourself towards the answer.

4. Nov 5, 2012

### haruspex

You could use cos(3θ) = 4cos3(θ) - 3cos(θ). Writing $θ=\pi/9$ gives you a cubic to solve.

5. Nov 5, 2012

### micromass

Staff Emeritus
Right. Thank you. I was thinking of the fact that the 9-gon wasn't constructible, but it is of course wrong to deduce from that that it can't be written with roots. It is clear that $\cos(\pi/n)$ is always algebraic, but when can it be written using radicals?? Since the Galois group of the n-th root of unity is abelian, it can always be expressed using radicals. Nice.