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A tricky trigonometric problem

  1. Mar 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A cubic equation is given as:
    ##x^{3} -(1+\cos \theta +\sin \theta)x^{2} +(\cos \theta \sin \theta +\cos \theta +\sin \theta)x-\sin \theta \cos \theta=0##

    Show that x=1 is a root of the equation for all values of θ and deduce that x-1 is a factor to the above equation.

    Hence, by factoring the cubic equation above, show that
    ##(x-1)[x^{2}-(\cos \theta +\sin \theta)x+\cos \theta \sin \theta]=0##

    and the roots are given by
    ##1, \cos \theta, \sin \theta##
    Write down the roots of the equation given that ##\theta=\frac{\pi}{3}##

    Find all values of ##\theta## in the range ##0<=\theta<2\pi## such that two of the three roots are equal.

    By considering ##\sin \theta -\cos \theta##, or otherwise, determine the greatest possible difference between the two roots, and find the values of ##\theta## for ##0<=\theta<2\pi## for which the two roots have the greatest difference.



    2. Relevant equations
    Factor theorem, trigonometric equations


    3. The attempt at a solution
    For the first part, I have plugged in x=1 and found that it is 0. Then I deduced that (x-1) is a factor.

    I factored the cubic above and factored the quadratic, and the roots are 1, ##\cos \theta## and ##\sin \theta##. Then I plugged in ##\theta=\frac{\pi}{3}## and found out the solutions are
    ##1, \frac{1}{2}, \frac{sqrt{3}}{2}##. Is it correct?

    Since the two roots are equal, therefore I set the following equations:
    ##\cos \theta =1##
    ##\sin \theta=1##
    ##\cos \theta=\sin \theta##

    The values were found out to be 0, ##\frac{\pi}{4}##, ##\frac{\pi}{2}## and ##\frac{5\pi}{4}##. Hopefully I did not make any mistakes here... :P

    For the last part, I need to consider ##\sin \theta-\cos \theta##. But how do I find the difference between the two roots?
     
  2. jcsd
  3. Mar 28, 2014 #2

    Simon Bridge

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    I cannot fault your method - did not check your arithmetic though.
    I suspect that the only roots they want to be the same are the sin and cos ones - so you have overkill.

    If one root is a and the other is b, then the difference between the roots is |a-b|.
    In your case the difference will depend on the angle.
     
  4. Mar 29, 2014 #3
    Could it be 2?
     
  5. Mar 29, 2014 #4

    Simon Bridge

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    Did you sketch out the graph of each root vs angle?
     
  6. Mar 29, 2014 #5
    Yup. One is sqrt(2), and another two is 2.
     
  7. Mar 29, 2014 #6

    ehild

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    No. there are three roots, 1, sinθ, cosθ, any pair of them can be equal.

    ehild
     
  8. Mar 29, 2014 #7

    ehild

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    I think, here you really need only the roots cosθ and sinθ. You are right, the greatest difference between them is √ 2.

    ehild
     
  9. Mar 29, 2014 #8
    Wasn't it should be 2? Since ##\sin \theta =1## and ##\cos \theta=1##?
     
  10. Mar 29, 2014 #9

    Simon Bridge

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    For what values of ##\theta## is ##|\sin\theta-\cos\theta | = 2##?
     
  11. Mar 29, 2014 #10

    ehild

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    sin2θ+cos2θ=1. Can both of them have magnitude 1?

    ehild
     
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