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Math Help - [SOLVED] Help sovling a simple equation

  1. #1
    brenoink
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    [SOLVED] Help sovling a simple equation

    sin(theta)(V2) - (V1)(d1) = (d2)cos(theta)

    I need to solve this for theta.

    I wasn't sure where to put this, so I put it here as it seems most general.

    This isn't for a class (it's for some code I'm working on), so I don't really know what section you'd want this under. Algebra 1, maybe?

    I feel sort of dumb for not being able to get this, as it LOOKS very simple, but I've tried every trig identity I know and I simply cannot simplify for theta.

    Any help would be much appreciated.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by brenoink View Post
    sin(theta)(V2) - (V1)(d1) = (d2)cos(theta)

    I need to solve this for theta.

    I wasn't sure where to put this, so I put it here as it seems most general.

    This isn't for a class (it's for some code I'm working on), so I don't really know what section you'd want this under. Algebra 1, maybe?

    I feel sort of dumb for not being able to get this, as it LOOKS very simple, but I've tried every trig identity I know and I simply cannot simplify for theta.

    Any help would be much appreciated.
    It's not as easy as it looks, so don't feel so bad. Here's how I would attack this:
    V_2~sin(\theta) - V_1d_1 = d_2~cos(\theta)

    Square both sides:
    V_2^2~sin^2(\theta) - 2V_1V_2d_1~sin(\theta) + V_1^2d_1^2 = d_2^2~cos^2(\theta)

    Now cos^2(\theta) = 1 - sin^2(\theta), so
    V_2^2~sin^2(\theta) - 2V_1V_2d_1~sin(\theta) + V_1^2d_1^2 = d_2^2~(1 - sin^2(\theta))

    Now to clean some stuff up.
    (d_2^2 + V_2^2)~sin^2(\theta) - 2V_1V_2d_1~sin(\theta) + (V_1^2d_1^2 - d_2^2) = 0

    This is a quadratic in sin(\theta). If you have a hard time seeing that (or dealing with it) let y = sin(\theta):
    (d_2^2 + V_2^2)y^2 - 2V_1V_2d_1y + (V_1^2d_1^2 - d_2^2) = 0

    So I get
    y = \frac{-(-2V_1V_2d_1) \pm \sqrt{(-2V_1V_2d_1)^2 - 4(d_2^2 + V_2^2)(V_1^2d_1^2 - d_2^2)}}{2(d_2^2 + V_2^2)}

    Now, if this isn't enough to have to clean up, you also have to make sure that you only take a solution that has a magnitude of one or less (because -1 \leq sin(\theta) \leq 1.) If you have numbers, now would be a good time to plug them in.

    I did a bit of work on this and got:
    sin(\theta) = \frac{V_1V_2d_1 \pm d_2\sqrt{V_2^2 - V_1^2d_1^2 + d_2^2 }}{2(d_2^2 + V_2^2)}
    I'd advise checking it to make sure I didn't make a mistake.

    (I feel this equation is relatively familiar. Where did it come from?)

    -Dan
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  3. #3
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    Trig

    First, you rearrange V_2\sin\theta-V_1d_1=d_2\cos\theta into
    V_2\sin\theta-d_2\cos\theta=V_1d_1 via simple algebra.
    Now, let R=\sqrt{V_2^2+d_2^2}, and let \phi be an angle such that tan\phi=\frac{d_2}{V_2} and such that \sin\phi has the same sign as d_2. Then, we see that V_2=R\cos\phi and d_2=R\sin\phi. Thus our equation becomes
    R\sin\theta\cos\phi-R\cos\theta\sin\phi=V_1d_1
    \sin(\theta-\phi)=\frac{V_1d_1}{R}
    \theta=\phi+\arcsin\frac{V_1d_1}{\sqrt{V_2^2+d_2^2  }}

    Note that if both V_2 and d_2 are non-negative, then 0\le\phi\le\frac{\pi}{2}, and \phi=\arctan\left(\frac{d_2}{V_2}\right) with the usual range on the inverse tangent.

    -Kevin C.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    First, you rearrange V_2\sin\theta-V_1d_1=d_2\cos\theta into
    V_2\sin\theta-d_2\cos\theta=V_1d_1 via simple algebra.
    Now, let R=\sqrt{V_2^2+d_2^2}, and let \phi be an angle such that tan\phi=\frac{d_2}{V_2} and such that \sin\phi has the same sign as d_2. Then, we see that V_2=R\cos\phi and d_2=R\sin\phi. Thus our equation becomes
    R\sin\theta\cos\phi-R\cos\theta\sin\phi=V_1d_1
    \sin(\theta-\phi)=\frac{V_1d_1}{R}
    \theta=\phi+\arcsin\frac{V_1d_1}{\sqrt{V_2^2+d_2^2  }}

    Note that if both V_2 and d_2 are non-negative, then 0\le\phi\le\frac{\pi}{2}, and \phi=\arctan\left(\frac{d_2}{V_2}\right) with the usual range on the inverse tangent.

    -Kevin C.
    Hmmmm...

    That's the second method tonight that I had totally forgotten about. Well, it's a good refresher anyway!

    -Dan
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