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How to get θ sin and cos are together in 1 equation

  1. Nov 15, 2011 #1
    b2cos2θ+2cosθ(-mb+bc2-bz-bcsinθ)+2sinθ(cm-cz)-c2sin2θ=p

    How to get θ in terms of b,c,m,z,p
    The equation i formed to calculate the dynamic
    PLs, anyone can help to solve this, i would much appreciate! im doing my FYP!
     
  2. jcsd
  3. Nov 15, 2011 #2

    HallsofIvy

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    Replace [itex]sin^2(\theta)[/itex] with [itex]1- cos^2(\theta)[/itex] and [itex]sin(\theta)[/itex] with [itex]\sqrt{1- cos^2(\theta)}[/itex]. That will give you an equation in [itex]cos(\theta)[/itex] only. However it will involve a square root so isolate that on one side of the equation and square both sides to get a fourth degree equation in [itex]cos(\theta)[/itex]. That is not going to give a simple answer!
     
  4. Nov 15, 2011 #3
    Thanks for ur precious reply! Somehow i still have to go through this question.. does it has any alternative method?
     
  5. Nov 15, 2011 #4

    phyzguy

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    HallsofIvy's approach is the right one. Do you know the values of b,c,m,z,p (or ranges)? The equation will be easily solved numerically if you do.
     
  6. Nov 15, 2011 #5
    b,c,m,z,p are the fixed constant, they will be known later. Somehow i cant get θ if i use HallsofIvy's approach because the existence of fourth, third, second, first degree of cos θ
     
  7. Nov 15, 2011 #6
    and also how if i want to use numerical method? how to use it ? Matlab?
     
  8. Nov 16, 2011 #7

    phyzguy

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    Personally, I use Mathematica, but Matlab will work. A simple Newton's method solver will find the root quickly once the coefficients are known.
     
  9. Nov 25, 2011 #8
    may i know how do i use numerical method to solve this trigo+algebraic equation? i need to equate the equation θ in terms of b,c,m,z,p

    Using Newton Raphson? (i donno how to use software, only know how to use handwritten)
     
  10. Nov 25, 2011 #9

    phyzguy

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    You could try Wolfram Alpha, which uses the Newton_Raphson method in the "FindRoot" function. For example, when I took the values, {b=1, c=2, z=3, m=4, p=5}, then using x for cos(theta), the equation becomes:

    (-5 - 6 x + x^2 - 4 (1 - x^2))^2 == (-4 + 4 x)^2 (1 - x^2)

    Wolfram Alpha, using the link below, finds x = -.45969. you can then find theta using the arccos function, or a calculator.

    http://www.wolframalpha.com/input/?...-4+(1-x^2))^2+==+(-4+4+x)^2(1-x^2),+{x,+0.5}]
     
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