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Math Help - Single equation, single unknown. How to solve?

  1. #1
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    Single equation, single unknown. How to solve?

    I have this equation:

    a+b+c*cos(\theta_4)=cos(d-\theta_4)

    I was wondering how do I solve for \theta_4?

    a,b,c,and d are known constants.
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  2. #2
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    Hello, s3n4te!

    It can be solved, but I'm sure you won't like it . . .


    Solve for \theta\!:\;\;a+b+c\cos(\theta) \:=\:\cos(d-\theta)

    We have: . \cos(d-\theta) - c\cos\theta \;=\;a+b

    . . \cos d\cos\theta + \sin d\sin\theta - c\cos\theta \;=\;a+b

    . . (\cos d - c)\cos\theta + \sin d\sin\theta \;=\;a+b .[1]


    Consider: . r^2 \:=\:(\cos d - c)^2 + (\sin d)^2 \;=\;\cos^2\!d - 2c\cos d + c^2 + \sin^2\!d \;=\;1 - 2c\cos d + c^2

    . . Then: . r \;=\;\sqrt{1-2c\cos d + c^2}


    Divide equation [1] by r\!:\;\;\frac{\cos d - c}{r}\cos\theta + \frac{\sin d}{r}\sin\theta \:=\:\frac{a+b}{r}


    Let \alpha be an angle such that: . \begin{Bmatrix}\sin\alpha &=& \dfrac{\cos d - c}{r} \\ \\[-3mm] \cos\alpha &=& \dfrac{\sin d}{r} \end{Bmatrix}
    . . Note that: . \tan\alpha \:=\:\frac{\cos d - c}{\sin d}


    Our equation becomes: . \sin\alpha\cos\theta + \cos\alpha\sin\theta \:=\:\frac{a+b}{r}

    Then: . \sin(\theta + \alpha) \;=\;\frac{a+b}{r} \quad\Longrightarrow\quad \theta + \alpha \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right)


    Therefore: . \theta \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right) - \alpha \quad\Longrightarrow\quad \boxed{\;\theta \;=\;\sin^{-1}\!\left(\frac{a+b}{r}\right) - \tan^{-1}\!\left(\frac{\cos d - c}{\sin d}\right)}

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  3. #3
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    Thank you very much for the tip!
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  4. #4
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    I have a question though.

    Can't you just make

    sin(\alpha) =cos(d)-c and cos(\alpha)=sin(d)

    Which would make \theta =sin^{-1}(a+b)-tan^{-1}(\frac{cos(d)-c}{sin(d)})

    What is the point of using r?
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