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Intersection Equation of two Trigonometric Graphs

  1. Aug 25, 2008 #1
    1. The problem statement, all variables and given/known data
    The graphs of f( ) = 2sin() - 1 (blue) , and g() = 3cos() + 2 (red)
    are shown below:
    [​IMG]
    [​IMG]

    What equation would have the intersection points of the graph its solutions?

    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    From using a calculator (set to degrees) I discovered the intersection points of the two graphs. They are:

    1: (112.61986, 0.84615385)
    and
    2: (180, -1)

    Therefore, to find an equation that will have these two points in its graph I did the following:

    2sin(x) - 1 = 3cos(x) + 2
    which eventually came to:

    2sin(x) - 3cos(x) = 3
    which brought me to:
    2sinxcosx - 3cosx=3
    cosx (2sins - 3) = 3

    so cosx = 3
    or sinx=3

    Needless to say, it is wrong.

    Can somebody tell me a strategy?

    Thanks in advance.
     
  2. jcsd
  3. Aug 25, 2008 #2

    dlgoff

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    "What equation would have the intersection points of the graph its solutions?"
    Couldn't you just find the equation for a line through the two points? Also, shouldn't you be finding the two points without a calculator?
     
  4. Aug 25, 2008 #3
    2sin(x) =/= 2sin(x)cos(x) = sin(2x). You confused 2sin(x) with sin(2x).

    Try writing everything in terms of sin(x). Setting the equation to 0, factoring, and checking for extraneous roots should solve the problem.
     
  5. Aug 25, 2008 #4
    If I wanted to find the points algebraically without a calculator, would I do what I thought was the original answer?

    That is solve for:

    2 sinx - 1 = 3cosx + 2

    ?

    Thanks
     
  6. Aug 25, 2008 #5

    dlgoff

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    Yes. This will give you the x value of the points. Now substitute these into one of the equations to find the y value.
     
  7. Aug 25, 2008 #6
    Thanks a lot.

    But I can't get past:

    2sin(x) - 3cos(x)=3

    How do I solve for x? Am I supposed to multiply them by some number so they all become sin?
     
  8. Aug 25, 2008 #7

    dlgoff

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    Do you know of any relations between the sin and cos functions so you would be able to eliminate one of them in your equation?
     
  9. Aug 25, 2008 #8
    is sin^2x=1-cos^2x?
     
  10. Aug 25, 2008 #9
    Oh, no. I multiply both sides by cosx.
     
  11. Aug 25, 2008 #10

    dlgoff

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    No. 2sin^2(x)=1-cos(2x)
     
  12. Aug 25, 2008 #11
    oh sorry
     
  13. Aug 25, 2008 #12

    dlgoff

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    So what do you get when doing this?
     
  14. Aug 25, 2008 #13
    I think you mean sin2x = 1 - cos2x. And yes that is the (fundamental) relation that would solve this problem. You might want to isolate the cos(x) term in the equation before applying that though.
     
  15. Aug 25, 2008 #14
    sin2x - 3cos^2x = 3cos x

    In other words, multiplying by cos (x) doesnt seem to get the answer.


    If i multiply by cos(x) I will never get rid of the cos(x) and won't be able to solve using one trig ratio.
     
  16. Aug 25, 2008 #15
  17. Aug 25, 2008 #16

    dlgoff

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    I'm stuck too. My brain isn't thinking.
    Are you sure the original equations are written correct?
     
  18. Aug 25, 2008 #17
    Dang, you kind of isolated the cos(x) term. Why not keep it simple and manipulate it so that you have positive terms?

    2sin(x)-3 = 3cos(x).

    By the fundamental pythagorean identity, cos(x) = [tex]\sqrt{1-sin^2x}[/tex]. Do you see how this substitution will make everything in terms of sin(x)?
     
  19. Aug 25, 2008 #18
    y= 2sinx - 1

    and

    y= 3cosx + 2

    therefore

    2 sinx - 1 = 3cosx + 2
    2 sinx - 3 cosx = 2 + 1
    2sinx - 3cosx = 3


    BTW, I thank you very much for trying.
     
  20. Aug 25, 2008 #19

    dlgoff

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    Yep. That should do it. Thanks snipez90
     
  21. Aug 25, 2008 #20
    let's see:

    (2sinx - 3) (2sinx -3) = 9 (1-sin^2x)

    4sin^2 - 6sinx - 6sinx + 9 = 9 - 9sin^2x

    4 sin^2x - 12sinx + 9 = 9 - 9 sin^2x

    13sin^2x - 12sinx = 0

    sinx (13sinx - 12) = 0

    Like that?
     
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