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Find theta

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    29θ - 17.7cosθ = 0
    find θ

    3. The attempt at a solution
    is there a specific way to do this without a calculator?
     
  2. jcsd
  3. Nov 1, 2008 #2

    cepheid

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    one thing I can think of...differentiate both sides.

    edit: well, even then, if you want an actual numerical answer for theta, I guess you'd need a calculator, unless you can do inverse trig functions in your head or use a table.
     
  4. Nov 1, 2008 #3

    gabbagabbahey

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    Hmmm... I don't think that taking the derivative is especially helpful here... Taking the derivative of [itex]29 \theta -17.7Cos(\theta)=0[/itex] gives you the same thing as taking the derivative of [itex]29 \theta -17.7Cos(\theta)=65967575865.79999678+68768i[/itex]. Just because a value of [itex]\theta[/itex] satisfies the equation [itex]f'(\theta)=0[/itex] does not necessarily mean that the same value of [itex]\theta[/itex] satisfies the equation [itex]f(\theta)=0[/itex]. After all, the anti derivative is only unique up to a constant. (i.e. [itex]f'(\theta)=0 \Rightarrow f(\theta)=constant[/itex])

    One useful method for estimating the root of the equation does involve derivatives however. If you plug pi/6 into the equation [itex]f(\theta)=29 \theta -17.7Cos(\theta)[/itex] you will see that f(pi/6) is pretty close to zero. You could then taylor series expand the function about the point pi/6 and keep say the first 2 or 3 terms of the expansion, then find the root closest to pi/6 of the resulting polynomial.

    Other than that, I would just plug it into your calculator.
     
    Last edited: Nov 1, 2008
  5. Nov 1, 2008 #4
    ok sorry, i can use a calculator for finding sines and cosines but i just meant that i cant plug the equation directly and get the answer like that. only because my calculator isnt sophisticated enough for that haha
     
  6. Nov 1, 2008 #5

    gabbagabbahey

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    Yes, taylor expanding the function to second order should give you a quadratic which you can solve with the help of a basic calculator, and give you a good estimate for theta (in fact, in this case exanding to second order gives a value of theta that is accurate to 6 decimal places!)
     
  7. Nov 1, 2008 #6
    wow, did you know this from experience or something? i would never have thought of using taylor series for anything other than programming applications. Thanks for the help, I'm going to try this and see how it works out
     
  8. Nov 1, 2008 #7

    gabbagabbahey

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    We use taylor series approximations a lot in physics. They are very useful for getting approximate solutions for complicated functions.
     
  9. Nov 2, 2008 #8

    cepheid

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    Taylor series occurred to me as well, but I wasn't sure if it was a good suggestion or not. As for the differentiation thing...yeah, I was being an idiot. Thanks for pointing that out gabbagabbahey.
     
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