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Sin/cos graph

  1. Aug 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve each equation for 0<theta<2pi


    a) 2sin(theta) + sincos(theta)=0

    2. Relevant equations





    3. The attempt at a solution

    I dont understand what to do, I know what your supposed to do when given one variable, which is rearrange and solve. But how do i solve for both of the variables here?
     
  2. jcsd
  3. Aug 1, 2011 #2

    SammyS

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    I only see one variable, θ.

    I take it that you are to solve: 2sin(θ) + sin(cos(θ)) = 0, correct?
     
  4. Aug 1, 2011 #3
    I see two, one being sin , one being cos. How to solve? no examples in the book no homework questions, Im clueless.
     
  5. Aug 1, 2011 #4
    Those are terms, not variables. Variables are values that can change with input. Well, put that equation in another way:

    sin(cos([itex]\theta[/itex])=-2sin([itex]\theta[/itex])

    It would be wise if you knew the values of unit circle. Like how sin 90º=1, cos 45º=[itex]\frac{\sqrt{2}}{2}[/itex], etc.
     
  6. Aug 1, 2011 #5
    I know all of the values of the unit circle, I dont understand how sin*cos = -2sin*

    , like what values does that give you... does that just say sin*cos as one term is = to -2sin.. so i inverse sin get the degrees then do 360- the value..?
     
  7. Aug 1, 2011 #6

    SammyS

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    sin & cos are functions.

    Also, in this case, I presume that the variable θ is in units of radians.

    BTW: By any chance, is the equation to be solved really:
    2sin(θ) + sin(θ)*cos(θ) = 0 .​
     
  8. Aug 1, 2011 #7
    Probably not but theres no trig identities that can change anything there
     
  9. Aug 1, 2011 #8

    SammyS

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    The equation 2sin(θ) + sin(θ)*cos(θ) = 0 can be solved by standard methods.

    The equation 2sin(θ) + sin(cos(θ)) = 0 can be solved numerically or graphically.

    Graph y = 2 sin(x) and y = -sin(cos(x)) on the same set of axes: attachment.php?attachmentid=37691&d=1312236844.gif
     

    Attached Files:

  10. Aug 1, 2011 #9
    numerically tbh
     
  11. Aug 1, 2011 #10
    Anyway, since your not helping with that. Atleast you can clarify something for me.

    If im given a period on a cosine function, like 2pi/3

    How do i turn that into a horizontal translation?

    Apparently 2pi/3 is = to 3x.

    I dont get how though.

    Is it because a period is 2pi over k, and the period im given is 2pi/3.

    Simply cross multiply and divide?
     
  12. Aug 1, 2011 #11

    SammyS

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    This question makes no sense (tbh).

    How is cosine related to 2pi/3 and 3x ?
     
  13. Aug 2, 2011 #12

    uart

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    Actually you were getting a lot of help. You were also making it as difficult as possible for people to help you by not having the courtesy to clearly state your original problem, despite being asked about it several times.

    So please, could you clearly state which of these is the original problem.

    Is it : 2 sin(theta) + sin(cos(theta)) = 0

    Or is it : 2 sin(theta) + sin(theta) cos(theta) = 0
     
  14. Aug 2, 2011 #13
    Here is a Cos graph from a test

    http://i53.tinypic.com/swtikn.jpg

    I understand how to get the vertical translation, by doing max-min/2 and getting the amplitude. How do you get the other translations by looking at the graph?
     
  15. Aug 2, 2011 #14

    eumyang

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    This is a different problem than stated in the OP. Please start a new thread next time.

    Are you supposed to use cos x as the basic graph?
    Using the general form y = a cos (b(x - h)) + k, you find b by using
    [itex]period = \frac{2\pi}{|b|}[/itex],
    which it looks like you wrote (in red ink).

    In the basic graph of y = cos x, there is a maximum point at (0, 1). In the transformed graph, there is a maximum point at (0, 2). So there is no horizontal translation (the h). You can also see that since the graph "centers" around the x-axis, there is no vertical translation either (the k). So the answer is what you wrote (in red ink):
    [itex]y = 2\cos \left( \frac{x}{4} \right)[/itex]
     
  16. Aug 2, 2011 #15

    PeterO

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    I wonder if you noticed that the question asked you to state the transformations, not the translations.

    Your equation was correctly derived, the transformations were:

    The Amplitude had been doubled - increased by a factor of 2 [from 1 to 2]
    The Period had be increased by a factor of 4 [from 2pi to 8pi]

    The graph had no obvious translations
     
  17. Aug 3, 2011 #16

    eumyang

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    I don't know about you, I consider a translation (shift) as a type of transformation. To me, transformations can be:
    * stretches/shrinks
    * reflections
    * translations (shifts)
     
  18. Aug 3, 2011 #17

    PeterO

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    Translations may be transformations, but not all transformations are translations.

    When Nero said "I know how to get the vertical translation from max-min / 2" he was clearly referring to a dilation , not a translation, though had used the term translation; incorrectly.
     
  19. Aug 3, 2011 #18

    eumyang

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    I'm not disputing that. What I tried to say earlier was that when you said this:
    ... it sounded like to me that you thought that transformations and translations were different things. I can now see that you didn't think that way.

    It also sounded like that there cannot be any translations involved in this problem. That's not entirely true: we all know that if we make the basic graph y = sin x instead of y = cos x, then yes, there would be a horizontal translation.
     
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