Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two equations, two unknowns

  1. Sep 21, 2004 #1
    I'm in a mechanics\statics class and have run across a couple of problems that I cannot, for the life of me, remember how to solve. The problems end up solving for two unknowns with two equations and one of the equations has
    Constant = Constant Sin(theta) + another constant Cos(theta)
    and the other equation has sin(theta) and some other unknown.

    My statics teacher will help me set up the engineering parts of the problem, but refuses to help me with the math. I should know how to do this, but I can't find it in any of my notes. I appreciate any help anyone can provide for this.
    Thanks,
    Josh
     
  2. jcsd
  3. Sep 21, 2004 #2
    Don't both equations have to have both unknowns in them to solve them using a realation?
     
  4. Sep 21, 2004 #3
    Could you write out the actual 2 equations u have and the 2 unknowns?
     
  5. Sep 22, 2004 #4
    A = B * sin(theta) + C * cos(theta)

    you can find the value of `theta' from this equation. the idea is to represent the right-hand side as sin(theta + alpha), where alpha depends on B and C.

    -- Adil
     
  6. Sep 22, 2004 #5
    Allright,
    I was mixed up here when I asked for help. I only have 1 equation, but I still can't solve it. The Eq is -180 = 217.5*cos(theta) + 101.9*sin(theta). I don't know how to solve for theta in this type of equation. I see sadrul's post below, but I'm not sure how to apply it. Thanks again for the help.
    --Josh
     
  7. Sep 23, 2004 #6
    To know how to solve an equation of the form

    [tex]A\cos\theta + B\sin\theta = C[/tex]

    you must know two identities in trigonometry, namely

    [tex]\sin(A+B) = \sin A\cos B + \cos A\sin B[/tex]
    [tex]\cos(A+B) = \cos A\cos B - \sin A\sin B[/tex]

    Additionally, you should know that both sine and cosine functions oscillate between -1 and +1. They can of course, assume the values -1 and +1. With a bit of work, you can show that

    [tex]A\cos\theta + B\sin\theta = \sqrt{A^2 + B^2}\sin\((\theta + \delta)[/tex]

    where [tex]\delta = \sin^{-1}\frac{A}{\sqrt{A^2 + B^2}}[/tex]

    Let's leave this an exercise for you so that you are at home with these equations (which will frequently arise in physics, engineering and trigonometry).

    From the above description, it should be clear that the equation will have a solution if and only if

    [tex]-\sqrt{A^2 + B^2} \leq C \leq +\sqrt{A^2 + B^2}[/tex]

    When you rearrange the final equation to solve for theta, you will most likely (except in mathematics where a general solution is required normally) attempt to find the principal value of the argument. That will be easy as you can simply find the inverse function using either a calculator or tables.

    Hope that helps...

    Cheers
    Vivek
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?