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How to solve 12sinx-1.8cosx=10 without a graphing calculator?

  1. Jan 24, 2012 #1
    I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

    Thanks!

    Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2012 #2

    ehild

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    Both sinx and cosx can be expressed by tan(x/2)

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

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


    With these substitutions, you get a quadratic equation for tan(x/2).

    ehild
     
  4. Jan 24, 2012 #3

    vela

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    When you have a combination like ##A \sin\theta + B \cos\theta##, a useful trick is to write
    $$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$Now identify
    \begin{align*}
    \cos\phi &= \frac{A}{\sqrt{A^2+B^2}} \\
    \sin\phi &= \frac{B}{\sqrt{A^2+B^2}}
    \end{align*}then you have
    $$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}(\cos\phi \sin\theta + \sin\phi\cos\theta) = \sqrt{A^2+B^2}\sin(\theta+\phi)$$where ##\tan\phi = B/A##.

    In this particular problem, if you move the factor of 4 kN over to the righthand side, you have
    $$3 \sin \theta - 0.45 \cos \theta = 2.5$$and ##\tan\phi = -0.45/3##, from which you should recognize ɸ as being the angle the line connecting A to the point where the force acts makes with the horizontal.
     
  5. Jan 24, 2012 #4

    SammyS

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    What system of equations.

    I see none.
     
  6. Jan 24, 2012 #5

    ehild

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    It is in the title. 12sinx-1.8cosx=10:smile:

    ehild
     
  7. Jan 24, 2012 #6

    SammyS

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    Thanks e . :smile:

    Any important information in the title should also appear in the body of the message.

    Also, OP refers to "... this system of equations".
     
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