1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    http://img403.imageshack.us/img403/6991/38ah7.jpg [Broken]
    http://g.imageshack.us/g.php?h=403&i=38ah7.jpg [Broken]

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

    2. Relevant equations

    http://www.eformulae.com/images/trigonometry_001.gif [Broken]

    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.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 25, 2008 #2

    dlgoff

    User Avatar
    Science Advisor
    Gold Member

    "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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Gold Member

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

    dlgoff

    User Avatar
    Science Advisor
    Gold Member

    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
    Last edited by a moderator: May 3, 2017
  17. Aug 25, 2008 #16

    dlgoff

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Gold Member

    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?
     
  22. Aug 25, 2008 #21
    Yes, and that means sin(x) = 0 or 13sin(x) - 12 = 0. Now check both of those equations. Remember there is probably a loss of information from squaring both sides, so make sure you only find the solutions that work. For instance, the graphs obviously don't intersect for x = 0 (which is an extraneous solution, sin(0) = 0 but 0 - 3 =/= 3).
     
  23. Aug 25, 2008 #22
    but, if I use:

    13sinx - 12=0
    13sinx = 12
    sinx = 12/13

    This equals 67 degrees.

    But the points of intersection are: (112.62, 0.8461) and (180, -1)
     
  24. Aug 25, 2008 #23
    I did warn you about extraneous solutions didn't I? :). Anyways your inverse sin on the calculator probably only calculates the least valued solution for which the equation is true. Remember sin(x) = sin(pi-x) pi is equivalent to 180 degrees by a conversion factor.
     
  25. Aug 25, 2008 #24
    I don't follow the last part you are talking about though. Why did you had 0-3?
     
  26. Aug 25, 2008 #25
    OK, I know I'm not supposed to give full solutions but you've pretty much solved the problem yourself (well hopefully you understood the motivation behind converting everything to sin(x) and well you did do the algebra yourself) so I'll try to explain the actual solutions/roots part.

    Whatever our solutions are, they have to satisfy 2sin(x) -1 = 3cos(x) + 2. Everything we do has to go back to that. Now we have sin(x) = 0 and 13sin(x) -12 = 0. Examine the first case sin(x) = 0. Now if you're just starting out, you would examine just two cases, x = 0 degrees and x = 180 degrees. If x = 0 degrees, 2sin(0) - 1 = -1 and 3cos(0) + 2 = 5, so this does not satisfy the equation. If x = 180 degrees, 2sin(180) -1 = 0 -1 = -1 and 3cos(180) + 2 = -3 + 2 = -1. So the equation is satisfied.

    Now generalize the result, from the unit circle, all multiples of pi, or 180 degrees, satisfy sin(x) = 0 (verify this). But we have shown that 0, an even multiple of pi does not satisfy the original equation. Thus all even multiples of pi are extraneous solutions (remember, adding 2pi or 360 degrees will give us the same angle essentially). Similarly, we have shown that pi satisfies the equation. So we claim that odd multiples of pi work. Thus we can write our solution set (from sin(x) = 0) as x = 180n degrees (or again, pi*x where n = 1, 3, 5 ...

    Now try analyzing the second part (13sinx - 12) on your own.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook