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Solve for A

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Is there a way to solve this for a?
    [itex]\frac{r\sin a}{t - r\cos a} = \cos \theta[/itex]



    [itex]r\sin a=t\cos \theta - r\cos a \cos \theta[/itex]

    2. Relevant equations



    3. The attempt at a solution
    I am not that good with trig...this isn't a homework problem but since it is like one it belongs here. My attempt at a solution got me to that equation and now I am stuck.
     
  2. jcsd
  3. Dec 8, 2011 #2

    Mentallic

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    Sure, do you know the method of converting

    [tex]a\cos(\theta)+b\sin(\theta)=Rsin(\theta+\phi)[/tex]

    Where R and [itex]\phi[/itex] are some values that you need to determine. You can start by expanding the right hand side and then equating the like terms.
     
  4. Dec 8, 2011 #3
    No...I am not aware of that at all. Does this method have a name by which I could search it? Or could anyone suggest a good trig site or book? I know that under certain circumstances you can swap trig functions but I don't really know what these situations are...

    What do you mean by expanding the right hand side?

    The known data that I am working with are: r t [itex]\theta[/itex]
     
  5. Dec 8, 2011 #4

    SammyS

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    To use
    Mentallic's suggestion, manipulate your equation to get:

    [itex]\displaystyle\sin(a)+\cos(\theta) \cos(a)= \frac{t}{r}\cos(\theta)[/itex]

    Look at Linear Combinations in the List of Trig Identities in Wikipedia -- this Link -- for the following:

    [itex]\displaystyle a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot\sin(x+\varphi)[/itex]

    [itex]\displaystyle \varphi = \arctan \left(\frac{b}{a}\right) + \begin{cases}
    0 & \text{if }a \ge 0 \\
    \pi & \text{if }a \lt 0
    \end{cases}[/itex]



    .
     
  6. Dec 9, 2011 #5
    I apologize but I am still quite confused...but without linear combination I came to this:

    [itex] \sin(a) + \cos(\theta)\cos(a) = \frac{t}{r}\cos(\theta)[/itex]

    expand [itex] \cos(\theta)\cos(a)[/itex]

    [itex] \sin(a) + \frac{\cos(\theta-a) - \cos(\theta +a)}{2} = \frac{t}{r}\cos(\theta)[/itex]

    What is confusing is that if I then use angle sum and difference I get

    [itex]\frac{\cos(\theta - a) - \cos(\theta + a)}{2} = \frac{\cos(\theta) \cos(a) - \sin(\theta) \sin(a) - [\cos(\theta) \cos(a) + \sin(\theta)\sin(a)]}{2} = \frac{-2\sin(\theta)\sin(a)}{2} = -\sin(\theta)\sin(a)[/itex]

    Plugging that back in above gives:

    [itex] \sin(a) - \sin(\theta)\sin(a) = \frac{t}{r}\cos(\theta)[/itex]

    Factor out sin(a)

    [itex] \sin(a)(1 - \sin(\theta)) = \frac{t}{r}\cos(\theta)[/itex]
    [itex] \sin(a) = \frac{t\cos(\theta)}{r-r\sin(\theta)}[/itex]

    So

    [itex] \sin^-1(\frac{t\cos(\theta)}{r-r\sin(\theta)}) = a[/itex]

    Any errors there? Seems odd... to go from "Product-to-sum" and then use "Angle sum and difference identities" But if that is allowable then it should be all good?
     
  7. Dec 9, 2011 #6

    NascentOxygen

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    Nevertheless, it has given you plenty of practice with itex. :smile:
    and you found it to be equal to
    Can this identity be correct, I hear you ask? So why not try a couple of random values for theta and a and evaluate on your calculator? I tried. Sorry, these are not generally equal. (I guess you've made a mistake in there somewhere.)

    What Mentallic intended when he said to expand, was that you should expand the RHS to remove the brackets. But in the nick of time SammyS came to your rescue and supplied you with result. So I recommend that you discard your off-track sum and difference extravaganza, and try to make use of SammyS's formula.
     
    Last edited: Dec 9, 2011
  8. Dec 10, 2011 #7
    I don't understand what SammyS means, because I can't seem to get my equation into that form...

    For clarification, I am going to change my equation's "a" variable since there is an "a" in the linear combination formula.

    [itex]\sin(\alpha)+\cos(\theta)\cos(\alpha)=\frac{t}{r}\cos(\theta)[/itex]

    Unless it is like this:

    [itex]a\sin x + b\cos x = \sqrt{a^2+b^2}\cdot\sin(x+\varphi)[/itex]

    [itex] a = 1 , b = \cos \theta [/itex]

    So that:

    [itex] \frac{t}{r}\cos\theta = \sqrt{1 + \cos^2\theta}\cdot\sin(\alpha + \varphi) [/itex]?

    Continuing with that I get
    [itex] \frac{t\cos\theta}{r\sqrt{1+\cos^2\theta}} = \sin(\alpha +\varphi) [/itex]

    [itex] \arcsin(\frac{t\cos\theta}{r\sqrt{1+\cos^2\theta}}) = \alpha +\varphi [/itex]

    [itex] \arcsin(\frac{t\cos\theta}{r\sqrt{1+\cos^2\theta}}) - \varphi = \alpha[/itex] ??

    In my case, [itex] \alpha \geq 0 [/itex] so it becomes:

    [itex] \arcsin(\frac{t\cos\theta}{r\sqrt{1+\cos^2\theta}}) - \arctan(\cos\theta) = \alpha[/itex] ??

    Thanks, it certainly looks much better than typed out equations!
     
  9. Dec 10, 2011 #8

    Mentallic

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    Very nice :smile:

    And just to finish off by showing you how the result of the substitution we made came to be:

    [tex]a\cdot\cos\theta+b\cdot\sin\theta[/tex][tex]=R\sin\left(\theta+\phi\right)[/tex]

    [tex]RHS=R(\sin\theta\cos\phi+\cos\theta\sin\phi)[/tex]

    Since we want to solve for [itex]\theta[/itex], we can equate the coefficients of [itex]\cos\theta[/itex] from each side, and also for [itex]\sin\theta[/itex].

    [tex]a\cdot\cos\theta \equiv R\cos\theta\sin\phi[/tex]
    [tex]a = R\sin\phi[/tex]

    and

    [tex]b\cdot\sin\theta \equiv R\sin\theta\cos\phi[/tex]
    [tex]b=R\cos\phi[/tex]

    So now we have these two equations in 2 unknowns (R and [itex]\phi[/itex]) and can thus solve them simultaneously. We can solve for [itex]\phi[/itex] by dividing the first equation by the second to get

    [tex]\frac{a}{b}=\tan\phi[/tex]

    [tex]\phi=\tan^{-1}\frac{a}{b}[/tex]

    and plugging this back into the first or second equation, we get

    [tex]a=R\sin\left(\tan^{-1}\frac{a}{b}\right)[/tex]

    and by playing around with this and applying trig substitutions (I myself prefer using a right triangle), you can determine that

    [tex]\sin(\tan^{-1}x)=\frac{x}{\sqrt{x^2+1}}[/tex]

    And so

    [tex]R=\frac{a\sqrt{x^2+1}}{x}[/tex]
    [tex]=\frac{a\sqrt{\frac{a^2}{b^2}+1}}{\frac{a}{b}}[/tex]
    [tex]=\frac{\frac{a}{b}\sqrt{a^2+b^2}}{\frac{a}{b}}[/tex]
    [tex]=\sqrt{a^2+b^2}[/tex]

    as required.
     
  10. Dec 10, 2011 #9
    I appreciate the help, but the function is not giving me the outputs I wanted...probably a problem with the original equation. It -should- be undefined for [itex]t = \sqrt{2}, r = 1, \theta \geq 45[/itex] but it doesn't seem to be the case
     
  11. Dec 10, 2011 #10

    Mentallic

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    Why should it be undefined for those given values?

    [tex]\frac{\sin(a)}{\sqrt{2}-\cos(a)}[/tex] has a range of [-1,1] and only reaches the extremities when [itex]a=\pi/4,7\pi/4[/itex], while on the right side of the equation, [itex]\cos\theta[/itex] ranges from -1 to 1 also.
     
  12. Dec 10, 2011 #11
    I don't know how to put it mathematically, but from my goal with the equation, it should.

    It is an attempt to find the angle of a radius with 0x, that goes to the point of intersection between a line originating at a Point "t" from the origin, with a circle of radius "r" centered around the origin, with the line making an angle of [itex]\theta[/itex] with 0x. If the situation of [itex]t = \sqrt{2}, r = 1[/itex] the point is at the corner of a square which the circle would be inscribed in, giving the two legs of the triangle that enclose the angle lengths of [itex]\sqrt{2}, 1[/itex], and the side opposite the angle length 1. with this setup the greatest angle from 0x that still intercepts the circle is [itex]\frac{pi}{4}[/itex], as that tangent line points directly at the point that is "t" from the origin


    Let me try to explain better, consider a circle of radius "r" center around the origin O, there is a point P on the x-axis that is a distance "t" from O, a line goes from P, at an angle of [itex]\theta[/itex] with the x-axis, intercepting the circle at a Point Q, what is the angle(with respect to the x-axis) of a line going from O to Q

    So if the line doesn't intercept the circle, there is no Q. Also, [itex]\overline{OQ} = r[/itex] should be true.

    This is why I say it is probably a problem with my original equation.
     
  13. Dec 11, 2011 #12

    NascentOxygen

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    Oh well, on the positive side you learned some valuable things from this exercise. :smile: :smile:

    I had a look at your derivation. I make the RHS of your equation to be -tan theta, not cos theta.

    Start again .... :uhh:
     
  14. Dec 11, 2011 #13
    tan indeed...and now it works. heh. Thank you three for your assistance, as that was certainly the way to go. Final version is:

    [itex] \arcsin(\frac{t\tan\theta}{r\sqrt{1+\tan^2\theta}}) - \theta = \alpha[/itex]
     
  15. Dec 11, 2011 #14
    I guess that is better expressed as:

    [itex]\arcsin(\frac{t}{r}\tan\theta\cos\theta)-\theta = \alpha[/itex]

    Can it be further simplified?
     
  16. Dec 11, 2011 #15

    SammyS

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    arctan(cos(θ)) ≠ θ
     
  17. Dec 11, 2011 #16
    "b" from linear combination equation was changed to a tan(θ), as that was the error in my original equation, is that what you mean?
     
  18. Dec 11, 2011 #17
    [itex]\arcsin(\frac{t}{r}\sin\theta)-\theta = \alpha[/itex] even then yes?
     
  19. Dec 11, 2011 #18
    Sorry to go on about this...but The above is correct for what I was going for, I was able to find it from a geometric proof working backwards from that answer.

    Circle.jpg

    [itex]\overline{OQ} = r, \overline{OP} = t, \angle{OPS} = \theta, \angle{QOP} = \alpha, \angle{OQS} = \varphi, \angle{OQP} = \beta[/itex]

    [itex]r\sin\varphi = t\sin\theta [/itex]
    [itex]\varphi+\beta = \pi[/itex]
    [itex]\alpha+\beta+\theta = \pi[/itex]

    So by solving for [itex]\varphi[/itex] in the first equation, equating the last two and removing [itex]\beta[/itex], equate the resultant equations via [itex]\varphi[/itex] and solve for [itex]\alpha[/itex] you reach the above equation.
     
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