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Plz help a little tricky TRIG eqn or SOlve not sure

  1. Mar 3, 2005 #1
    The graphs of f(theta)=2sin(theta)-1 BLUE and g(theta)=3cos(theta)+2 RED
    are shown at this link

    http://ca.pg.photos.yahoo.com/ph/sikandar1984/detail?.dir=9181&.dnm=ebc3.jpg&.src=ph

    The original QUESTION: What equation would have the intersection points of the graphs its solutions?

    The above didnt make sense to me so the teacher reworded it
    PRESENT QUESTION: If you were to find the intersection of both graphs, what equations would you use?

    Hint is that these two graphs intersect when both equations are equal.

    Im not sure what the question is asking does it want an equation or a solution?

    so far ive put both equations equal to eachother and after simplifying ive got up to 4sin^2(x)-12sin(x)+9=9(1-sin^2(x)) i dont know whats next or if this is what the question is asking for plz help! :smile:
     
  2. jcsd
  3. Mar 3, 2005 #2

    dextercioby

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    Besides the obvious
    [tex] 3\cos\theta +2=2\sin\theta -1 [/tex],i don't see another one...

    Daniel.

    P.S.They're not asking for the solutions,just the equation.
     
  4. Mar 3, 2005 #3
    hmmm anyone else any ideas? Everytime i asked online tutors for help at ilc.org they said somethin different..im goin nuts :cry:
     
  5. Mar 3, 2005 #4

    dextercioby

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    Trust me,no one else could understand this statement otherwise:"What equation would have the intersection points of the graphs its solutions???"I think the degree o clarity is more than satisfactory.

    Daniel.
     
  6. Mar 3, 2005 #5

    James R

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    I agree with dextercioby.
     
  7. Mar 3, 2005 #6

    dextercioby

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    BTW,what's 'ilc.org'...?Illicite mathematical advice...?dot org...

    :wink:

    Daniel.
     
  8. Mar 3, 2005 #7

    HallsofIvy

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    What exactly are you asking?

    Your teacher asks "what equation... ?" and you say you don't understand so the teach rewords it and asks "if .... what equations ...?" and you say you don't understand if they are asking for "equations or solutions". Then dextercioby says "They're not asking for the solutions,just the equation."

    How about just assuming that people actually mean what they say?
     
  9. Mar 4, 2005 #8
    no no of course I believe what ur saying about the answer being an equation but there is this online site where tutors help students after school, and for some reason everytime I asked them the same question they started solving instead of finding an equation but I never got a chance to ask them why they were solving, they would say too many ppl in line have to move on, thats why im so confused I guess im gonna go with what dextercioby said and see what marks I get good luck to me lol :redface:
     
  10. Mar 4, 2005 #9

    dextercioby

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    If you presented them with the text of the problem in the EXACT form which u posted on PF,then they're a bunch of incompetents...

    Daniel.
     
  11. Mar 4, 2005 #10
    [tex] 3\cos\theta +2=2\sin\theta -1 [/tex]

    would it be better to make the trig functions the same and then bring everything over to the left side collecting like terms and making the right hand side of the equation 0?
     
  12. Mar 4, 2005 #11

    dextercioby

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    Of course you can do that...:bugeye:It's called the "canonical" form of an equation:"function of (unknown)=0"...

    Daniel.
     
  13. Mar 4, 2005 #12
    [tex] 2 \sin\theta-1=3\cos \theta+2 [/tex]

    [tex] 2\sin\theta- 3 \cos\theta-3=0 [/tex]

    Let sin theta = x

    2x-3sqrt(1-x^2) -3=0

    (2x-3)= 3sqrt(1-x^2)

    [tex]4x^2-12x+9=9(1-x^2)[/tex]

    [tex] 13x^2-12x=0 [/tex]
    [tex] 13\sin\theta^2-12\sin\theta [/tex] Is this an ok final answer?
     
    Last edited: Mar 4, 2005
  14. Mar 4, 2005 #13

    dextercioby

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    Unless you put it like this
    [tex]13(\sin\theta)^{2}-12\sin\theta=0 [/tex]
    ,no.

    Daniel.
     
  15. Mar 5, 2005 #14

    HallsofIvy

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    Did you understand dextecioby's point?

    1: [itex]sin \theta^2[/itex] is NOT the same as [itex]sin^2\theta= (sin\theta)^2[/itex]

    2: Even [itex] 13 sin^2\theta- 12 sin\theta[/itex] is not correct because it is NOT an equation!
     
  16. Mar 5, 2005 #15

    Curious3141

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    Can't you just leave the answer as [tex]2\sin\theta - 3\cos\theta = 3[/tex] ? After all, they only wanted an equation, and that's an equation.

    At any rate, Aisha, the final equation you ended up with (even with dextercioby's correction) is WRONG !

    Squaring an equation to find the solutions is dangerous unless you know exactly what you're doing. In this case, you ended up introducing redundant roots and might miss the ones that you should actually have found. If you solve the equation you ended up with and get [tex]\sin \theta = 0[/tex] OR [tex]\sin \theta = \frac{12}{13}[/tex] and use those to solve for [tex]\theta[/tex], you'd have two extra roots that do not satisfy the original equation.

    Try solving it like this :

    [tex]2\sin\theta - 3\cos\theta = 3[/tex]

    [tex]\sqrt{13}\sin{(\theta - \arctan{\frac{3}{2}})} = 3[/tex]

    That reduction is based on addition formulae. It's a slightly more advanced technique you may not be familiar with, nevertheless it's the only correct one to use here. Note that the arctan(3/2) is only the 1st quadrant value and we're only taking the positive root of 13.

    Solving you get :

    [tex]\theta = \arcsin{\frac{3}{\sqrt{13}}} + \arctan{\frac{3}{2}}[/tex] where the arcsin can be in the first and second quadrants, whereas the arctangent is only the first quadrant value.

    Further, you'll find with a (2,3, [itex]\sqrt{13}[/itex]) right triangle that the principal values of those two angles come from the same right triangle. Therefore, the solution can be reduced to :

    [tex]\theta = \pi[/tex] (or 180 degrees)

    OR [tex]\theta = 2\arctan{\frac{3}{2}}[/tex], which is around 112.6 degrees.
     
    Last edited: Mar 5, 2005
  17. Mar 5, 2005 #16
    nice! that was well done, Curious3141!
     
  18. Mar 5, 2005 #17

    Curious3141

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    Thanks. :biggrin:
     
  19. Mar 5, 2005 #18

    xanthym

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    .
    Curious --

    Your point is well-taken. However, there are practical trade-offs in solving those problems which require "squaring terms" regarding whether your "more advanced, more complex" approach is the "best practical" method or whether the following rule would yield the same results:
    "If solving an equation ever requires squaring terms, extraneous roots may be generated, and all resulting solutions must be checked with the original equation, and any non-correct solutions DISCARDED."
    On a practical basis, the above rule may be more appropriate for most cases encountered in this forum.



    ~~
     
    Last edited: Mar 5, 2005
  20. Mar 5, 2005 #19

    Curious3141

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    There are cases when squaring is unavoidable, such as in surd equations, in which case, checking for redundant roots becomes necessary.

    This is not one of those cases. The problem can be solved more elegantly (and truly simply, IMHO) using a simple observation about the sums of sines and cosines. If such a method exists, it should be used.
     
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