Finding cos in radians without calculator? Help?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
4 replies · 16K views
mathFun
Messages
7
Reaction score
0
Hey guys this question is going to 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?
 
Mathematics news on Phys.org
This wiki article might be useful: http://en.wikipedia.org/wiki/Exact_trigonometric_constants
It shows how to find things like [itex]\cos(\pi/5)[/itex] using geometry.

Note that it is not always possible to find explicit value of cosines. For exampe, the innocent looking [itex]\cos(\pi/9)[/itex] cannot be explicitely found using elementary operations and roots (unless I'm being stupid right now).
 
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.
 
micromass said:
the innocent looking [itex]\cos(\pi/9)[/itex] cannot be explicitly found using elementary operations and roots
You could use cos(3θ) = 4cos3(θ) - 3cos(θ). Writing [itex]θ=\pi/9[/itex] gives you a cubic to solve.
 
haruspex said:
You could use cos(3θ) = 4cos3(θ) - 3cos(θ). Writing [itex]θ=\pi/9[/itex] gives you a cubic to solve.

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 [itex]\cos(\pi/n)[/itex] 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.