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Angle panic

  1. Jul 8, 2012 #1
    Angle panic!!

    When we have a relation like, for instance,

    [itex] f(\theta) + g(\theta) = constant [/itex] where [itex]\theta[/itex] is an angle, does it hold for any angle such that we can do [itex] f(0 \deg) + g (0 \deg) = 0 [/itex] and we would obtain an universal result? I mean, imagine [itex] f(\theta, x) = x \sin \theta [/itex], then

    [itex] x = - \frac{g (0 \deg)}{\sin 0\deg} = - \frac{g (45 \deg)}{\sin 45\deg} [/itex]
     
  2. jcsd
  3. Jul 8, 2012 #2
    Re: Angle panic!!



    I've read the above 4 times (the last two rather slowly and carefully) and I still cannot understand what it means...

    One thing is sure, though: [itex]\sin 0 = 0\\,[/itex] so it cannot appear in the denominator.

    DonAntonio
     
  4. Jul 8, 2012 #3
    Re: Angle panic!!

    Sorry, bad choice on the sine...

    But imagine that you have the following equation:

    [itex] x \cos \theta - y(\theta) \cos^2\theta = 0 [/itex]

    I wish to find a solution for x. What I want to know is if it is equivalent:

    [itex] x = y (0 \deg) = \frac{y (45 \deg) \frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{ y (a \deg) \cos^2 a}{ \cos a} [/itex].

    Because, in my mind, it should be! If it holds for an unknown [itex] \theta [/itex], it should hold for any [itex] \theta [/itex]!
     
  5. Jul 8, 2012 #4
    Re: Angle panic!!

    That sine actually made me think on something, even if it holds, the equation would not be valid for that particular value of [itex] \theta [/itex].
     
  6. Jul 8, 2012 #5

    HallsofIvy

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    Re: Angle panic!!

    Then you can simply write [itex]x= y(\theta)cos^2(\theta)/cos(\theta)= ycos(\theta)[/itex] as long as [itex]cos(\theta)[/itex] is not 0- that is if [itex]\theta[/itex] is not an odd multiple of [itex]\pi/2[/itex].

    Again, it is not clear what you mean. What do you mean by "holds for an unknown"?
    [itex]x^3= 3[/itex] for some "unknown" value but does not hold for all x.
     
  7. Jul 8, 2012 #6
    Re: Angle panic!!

    I don't know if you will understand... but I want to know if any value for the angle, as long as it has a finite result, can be used to solve an equation with unknown angles.

    How would you solve this, for instance:

    [itex] x \sin \theta + \sqrt{x \cos \theta} = 0 [/itex] ?

    Could you assume any value for [itex] \theta [/itex] and the resulting x would be the same?
     
  8. Jul 10, 2012 #7

    haruspex

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    Re: Angle panic!!

    Certainly not. The above equation makes x and θ functions of each other. x = 0 is always a solution, and otherwise it can be simplified to [itex] x = cosec(\theta) cot(\theta) [/itex]
     
  9. Jul 10, 2012 #8
    Re: Angle panic!!

    I'm just writing equations for the sake of it... I'm not trying to solve them...

    And would this make sense: If I would make an average of the equation and solve for x, like this:

    Suppose you have: [itex] x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta = 0 [/itex]

    then to find a solution to x could you average the above equation on [itex] \theta [/itex] (NOTE: The above equation is just for illustrative purposes... I don't need help solving it!)

    [itex] \frac{1}{2 \pi} \int_0^{2 \pi} x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta d\theta = 0 [/itex].
     
  10. Jul 10, 2012 #9

    haruspex

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    Re: Angle panic!!

    I'm not at all sure I understand what you're trying to do.
    If you have an equation [itex] x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta = 0 [/itex], what do you mean by "finding a solution for x"? Normally the meaning would be to find a function of θ which yields the value of x for any given θ. Or sometimes there might be a specific θ that you are interested in and only want the value of x that goes with that. On yet other occasions (usually re Diophantine equations) you just want any pair x and θ that satisfies the equation.
    There may be yet other situations where you want to know the average value of x as θ goes through some range (you would need specify the weight given to each part of that range), but this would not be described as "finding a solution for x".
     
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