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Simple trig problem

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    I feel like I this problem shouldn't be that hard, but I cant figure out how to evaluate this equation

    cosx - cosx*sinx = 1/3

    2. Relevant equations
    I don't know what to use for this, none of the trig identities seem to help

    3. The attempt at a solution
    I tried substituting ## sqrt(1-cos^2x )## for sinx , but I just ended up with something way more complicated. I know this shouldn't be this hard because I'm not even in a trig class, Im taking statics. It wouldn't make sense for them to give me a problem that requires more time with the trig than with the statics.

    Thank You
     
  2. jcsd
  3. Oct 4, 2015 #2

    berkeman

    User Avatar

    Staff: Mentor

    WolframAlpha.com shows solutions to this, but it doesn't seem to show a simplified trig form...
     
  4. Oct 4, 2015 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    This is simple to solve numerically, but not at all easy to solve "analytically" (i.e., using formulas). For what it's worth, Maple 11 obtains the following "analytic" solution:
    x =
    arctan((-1+3*(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2))^3)/(1/6*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+1/6*((-(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)*(162+6*681^(1/2))^(2/3)-12*(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2)+36*(162+6*681^(1/2))^(1/3))/(162+6*681^(1/2))^(1/3)/(((162+6*681^(1/2))^(2/3)+12)/(162+6*681^(1/2))^(1/3))^(1/2))^(1/2)))
    Basically, this has the form
    [tex] x = \arctan\left( \frac{-1 + 3z^3}{z} \right), [/tex]
    where ##z## is a root of the equation ##9z^4 -6z + 1 = 0##. Numerically, ##x \doteq 0.6284918352##. Of course, there are other solutions outside the region ##0 < x < \pi##.
     
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