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Finding cos in radians without calculator? Help?

  1. Nov 5, 2012 #1
    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. jcsd
  3. Nov 5, 2012 #2
    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).
  4. Nov 5, 2012 #3


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    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.
  5. Nov 5, 2012 #4


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    You could use cos(3θ) = 4cos3(θ) - 3cos(θ). Writing [itex]θ=\pi/9[/itex] gives you a cubic to solve.
  6. Nov 5, 2012 #5
    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.
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