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Quick trigonometric function problem

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Sketch the curve of y=sin(x)+sin(2x)

    I got all the derivatives, and the roots N(k*∏|0), but now I'm stuck with the maxima/minima


    2. Relevant equations
    First derivative y'=cos(x)+2cos(2x)


    3. The attempt at a solution

    I set cos(x)+2cos(2x)=0 but now I'm stuck. I just don't know how to solve this oO
    Graph-plotters and the solution book spit out -2,57 -0,94 and so on, but how do I get this solution?
     
  2. jcsd
  3. Oct 29, 2013 #2

    Dick

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    Use a double angle formula to express cos(2x) in terms of cos(x).
     
  4. Oct 29, 2013 #3
    So cos(2x) = 2cos^2(x)-1?

    When I calculate that through I get cos(x)+4cos(x) = 1

    What now? Sorry, I'm really bad at trigonometric identities :(
     
  5. Oct 29, 2013 #4

    Mark44

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    What does ∏|0 mean?
     
  6. Oct 29, 2013 #5

    Dick

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    What happened to the cos(x)^2? You should get a quadratic equation in cos(x). Use that to find the possible values of cos(x).
     
  7. Oct 29, 2013 #6

    Student100

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    You made a mistake with your multiplication here, look at what the 2 should be multiplying, you also lost a power.
     
  8. Oct 29, 2013 #7
    Sorry, I made a typo, I meant I get cos(x)+4*cos^2(x) = 1
    I don't know what to do with this equation :(

    @Mark: It means x=k*pi and y=0
     
  9. Oct 29, 2013 #8

    Student100

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    Still wrong. 2[cos(2x)]
     
  10. Oct 29, 2013 #9
    How can I change cos(x)+2cos(2x) to 2[cos(2x)]? oO
    I don't get it :(
     
  11. Oct 29, 2013 #10

    Student100

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    No no, take a realllll close look at your algebra after you switch the 2[cos(2x)] with 2[2cos2(x)-1].

    Since you want to find the roots, what kind of equation does this now produce? What can you do with this equation to make your life easier? What has to be done to the equation to find the roots? Hint: Don't move the constant like you've been doing and maybe it'll all make more sense.
     
  12. Oct 29, 2013 #11
    Oh okay, so I get cos(x)+4cos^2(x)-2 = 0

    And now? Oo
     
  13. Oct 29, 2013 #12

    Student100

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    Haha good job.

    Now what type of equation is that?
     
  14. Oct 29, 2013 #13
    It's a quadratic equation, but can I use the formula to solve those for cos(x) too?
    Sorry, I'm so rusty in algebra and math, haven't been doing any for 3 years now *sigh*
     
  15. Oct 29, 2013 #14

    Student100

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    Don't feel bad, I'm still pretty bad at math.

    You certainly can. You could use u substitution to put it in terms of u if it's confusing for you:
    u = cos(x)
    4u2 + u -2.

    Or you could also just plug it into the quadratic formula as is.

    You just have to remember that cos(x) oscillates, so you have to put the anwser in terms of 2[itex]\pi[/itex]n. Unless your question states a closed interval. Also, you have to use the inverse trig function.

    You can brush up on algebra here: http://www.purplemath.com/modules/index.htm
     
    Last edited: Oct 29, 2013
  16. Oct 29, 2013 #15
    So the answer would be cos(x) = -1(+-)sqrt(33)/8 ?

    This gives me the solutions from above :) Thank you :)

    How do I convert this into 2pi*n though?
     
  17. Oct 29, 2013 #16

    Student100

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    Does the problem state an interval to draw the original function on? It's probably assumed you'd only have to draw from [itex] (-2\pi , 2\pi) [/itex] So it isn't important for this problem most likely, but it's nice to know:

    [itex] 2\pi n \pm \arccos( \frac{-1 \pm \sqrt(33)} {8})[/itex]

    This would give you the relative extrema for as long as you'd want to look at the original function.
     
    Last edited: Oct 29, 2013
  18. Oct 29, 2013 #17
    Thank you :)
     
  19. Oct 29, 2013 #18

    Mark44

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    What you wrote is something different from what you mean.
    This is what you wrote:
    $$cos(x) = - 1 \pm \frac{\sqrt{33}}{8}$$

    What you meant was this:
    $$cos(x) = \frac{-1 \pm \sqrt{33}}{8}$$

    If you don't use LaTeX to write this, you need parentheses around the entire numerator, like so:
    cos(x) = (-1 ± √33)/8

    Apparently what you wrote confused Student100, who is not working with the correct value.

    Student100, I don't believe this is correct. It looks like you are saying cos-1(A + B) = cos-1(A) + cos-1(B). That is NOT true.
     
    Last edited: Oct 29, 2013
  20. Oct 29, 2013 #19

    Student100

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    Your right Mark, I was trying to save time by putting everything in one expression. I could see how that might be confusing.

    Quick question:

    Which root is outside the domain? I thought I did a quick check on the calculator and both were valid. I may be wrong.
     
  21. Oct 29, 2013 #20

    Mark44

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    Not just confusing, but wrong, if my understanding of what you did is correct. Is this what you did? cos-1(A + B) = cos-1(A) + cos-1(B)?
    Neither is outside the domain. I got fixated on the OP's solution of -1 ± (√33)/8 and this threw me off. The correctly written values, (-1 ± √33)/8, are both within the interval [-1, 1].
     
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