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Trigonometric Circle : f(x) = cos(2x+1)

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    I have the function f(x) = cos(2x+1)
    I need to resolve f(x) = 0.6 when -PI <= x <= PI

    then put my answers in a Trigonometric Circle

    2. Relevant equations


    3. The attempt at a solution
    First I do this:
    cos(2x+1) = 0.6 and use my TI calculator to solve for X and I find the following solutions
    x1 = -0.964 x2 = -0.036 x3 = 2.178 x4 = 3.105

    now to put those in a Trigonometric Circle am I supposed to use my x in the f(x) function and do like
    f(x1) = cos(2x1+1) = cos(-0.964) = 0.599436
    or directly from x1-x4 to my Trigonometric Circle?

    thank you very much

    EDIT: I dont understand because my cos in my Trigonometric Circle goes from 1 to -1 how am I supposed to put x3 = 2.178 x4 = 3.105 :/
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2
    Great work, just to remind that, in geometry and also by definition, cos and trig function are functions of angles, :)
    Edit: Don't forget that you find a solution for 2x+, get the x out of that and plot them,
     
    Last edited: Apr 8, 2015
  4. Apr 8, 2015 #3

    SammyS

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    Well, you solved for x, didn't you?

    And f(x1) ≈ 0.6 . Right?

    x1, x2, x3, x4, don't go directly on your trig. circle.

    If you really want them on your trig. circle, multiply each of them by 2, then add 1, then find that number in radians on the circle. Each of those should be at x ≅ 0.6 , assuming that the x-coordinate on your circle corresponds to cosine .
     
  5. Apr 8, 2015 #4
    x1 = -0.964 x2 = -0.036 x3 = 2.178 x4 = 3.105

    x1 = (-0.964*2)+1 = -0.928
    x2= (-0.036*2)+1 = 0.928
    x2= (2.178*2)+1 = 5.356
    x3= (3.105*2)+1 = 7.21

    I dont really understand why we do *2+1 and how to find the result in radian
     
  6. Apr 8, 2015 #5

    BvU

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    Your calculator doesn't help you to understand what you are doing when you solve a tigonometric equation.

    If you want to solve ##\cos(2x+1) = 0.6##, you find an angle that has a cosine of 0.6.

    Now the calculator comes in handy: if ##\phi = 0.9273## radians (or 53.13 degrees), then ##\cos\phi=0.6##.

    So now you can write ##\cos(2x+1) = \cos\phi##, and this means that your solutions satisfy

    ##2x+1 = \phi + 2n\pi## or (because ##\cos(\theta) = \cos(-\theta)## )
    ##2x+1 = -\phi + 2n\pi##
     
  7. Apr 8, 2015 #6
    ϕ=0.9273 \phi = 0.9273 radians (or 53.13 degrees), then cosϕ=0.6 \cos\phi=0.6 .
    ϕ=0.9273 \phi = -0.9273 radians (or -53.13 degrees), then cosϕ=0.6 \cos\phi=0.6 .
    ϕ=0.9273 \phi = 5.356 radians (or 306.876 degrees), then cosϕ=0.6 \cos\phi=0.6 .
    ϕ=0.9273 \phi = 7.21 radians (or 413.103 degrees), then cosϕ=0.6 \cos\phi=0.6 . // Does that make sense that its more than 360 degree?
     
  8. Apr 8, 2015 #7
    I still need help please :(
     
  9. Apr 8, 2015 #8

    SammyS

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    x1 is -0.964 , it's not -0.928, etc. You only multiply by 2, then add 1 to use the trig. circle.

    You are solving ##\ \cos(2x+1)=0.6\ ##. So that to check any of your solutions for ##\ x\,,\ ## you need plug them in which amounts to multiplying your solution by 2, then adding 1 to that. Taking the cosine of this quantity should give of approximately 0.6 .

    In other words, since your answers are in radians, multiplying by 2 and then adding should result in an angle, which should correspond to a cosine of 0.6 on your trig. circle.
     
  10. Apr 9, 2015 #9

    SammyS

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    You seemed rather impatient for a response. Haven't heard back from you. Maybe that means you now are satisfied with your understanding or with your solution.


    Just a quick remark regarding Post #3 as a response to your Original Post.

    I should have mentioned that your solutions were a complete set and all correct, at least to the nearest 0.001 radian.

    I was simply trying to answer your questions - which showed much confusion - regarding how the solutions are related to the Trigonometric Circle.

    I apologize if I mislead you regarding the correctness of those solutions.

    SammyS
     
  11. Apr 9, 2015 #10

    BvU

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    From this I conclude you worked out ##\pm\phi + 2n\pi## with n = 0 and n = 1. You found two distinct angles for which the cosine is 0.6

    There is no problem for going over ##2\pi##, but adding ##2\pi## does not give a new solution fo ##\phi##: 413.13 degrees is the same as 53.13 degrees and 308.87 is the same as -53.13.

    However, there are more than 2 distinct solutions for ##x##, because of the factor 2.
    And the exercise asks you to place the answers for ##x## on the unit circle.

    So I would find it logical if you would first solve
    ##2x+1 =\pm \phi + 2n\pi\ \ ## for ##x##​
    and then look for the unique answers in a range of ##2\pi## wide: ##[0,2\pi]## or ##[-\pi,\pi]##

    The equation ##2x+1 =\pm \phi + 2n\pi\ \Rightarrow\ x = {\pm \phi - 1 \over 2} + n\pi\ ## gives you 2 solutions in ##[0,2\pi]## for the plus sign (n= 0 and n = 1), and also 2 for the minus sign (n = 1 and n = 2).

    I must say that the instruction to place the answers on the unit circle doesn't make much sense to me, but perhaps I miss something didactically relevant there.
     
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