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Trig functions periodicity

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Which one is not periodic?
    (a)|sin 3x|+sin2x
    (b)cos[itex]\sqrt{x}[/itex]+cos2x
    (c)cos 4x + tan2x
    (d)cos 2x+sin x


    2. Relevant equations



    3. The attempt at a solution

    I don't understand how to show whether the functions are periodic or not? :confused:
     
  2. jcsd
  3. Aug 10, 2011 #2

    SammyS

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    Did you try graphing each function? Also, graph each term of each function.
     
  4. Aug 10, 2011 #3
    I don't want to use graphs for this question since i am not comfortable in adding them.
    Any other method? :smile:
     
  5. Aug 10, 2011 #4
    Look at the arguments of the trig functions; are any different than the others?
     
  6. Aug 10, 2011 #5

    SammyS

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    What's the period of sin(3x)? What's the period of |sin(3x)| ?

    What's the period of sin2(x) ?

    Furthermore, regarding my previous post:
    Even if you're not comfortable adding graphs, graphing each term for each problem may help you.​
     
  7. Aug 10, 2011 #6
    For trigonometric functions, the period is generally expressed as a constant coefficient of the independent variable. Look through your choices for a term that does not contain a constant coefficient of the independent variable. Graph it to confirm that it is not periodic.
     
    Last edited: Aug 10, 2011
  8. Aug 10, 2011 #7

    NascentOxygen

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    Not comfortable with graphing? Then that's precisely why you should do some graphing.....and keep at it until you become perfectly comfortable with it. :smile:

    That's the secret to becoming more proficient at graphing.

    Here, try this: http://fooplot.com/"
    Type a different equation in each colour box on the right.

    I'll help you express your first equation: abs(sin(3x))+(sin(x))^2
     
    Last edited by a moderator: Apr 26, 2017
  9. Aug 11, 2011 #8
    What do you mean by "arguments"?:confused:

    How would i graph sin(3x)? And the period of sin2x is [itex]\pi[/itex].

    Sorry, i didn't get you. Please elaborate. :smile:
     
  10. Aug 11, 2011 #9
    The argument of sin(2x+5) is 2x+5, for cos(x+y) it's x+y, etc.

    The 3 in sin(3x) affects the graph basically in only one way. What's the period of sin(3x) compared to sin(x)?
    And it's correct, but how do you know the period of sin2x is [itex]\pi[/itex]?
     
  11. Aug 11, 2011 #10
    Using wolfram alpha i found that the period of sin(3x) is [itex]\frac{2\pi}{3}[/itex] and sin2 is [itex]\pi[/itex].
     
  12. Aug 11, 2011 #11

    SammyS

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    How about |sin(3x)| ?
     
  13. Aug 11, 2011 #12

    I like Serena

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    Hi Pranav-Arora! :smile:

    Hmm, the sine and cosine functions have a period of 2pi.
    This means sin(x) = sin(x+2pi) = sin(x+4pi) = sin(x+6pi) = ...

    So sin(3x) = sin(3x+2pi) = sin(3x+4pi) = ...

    What period does this imply for x?

    Similarly sin2x=(1 - cos(2x))/2
    From this you can deduce its period...
     
  14. Aug 12, 2011 #13
    Hi I Like Serena! :smile:

    If i take this case: sin(3x)=sin(3x+2pi), then 3x gets cancelled and i am left with 0=2pi. :(

    How would i deduce the period of sin2x using the equation given by you? :confused:
     
  15. Aug 12, 2011 #14

    PeterO

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    Post #13 makes it look like you don't recognise what the functions are???
     
  16. Aug 12, 2011 #15
    I know what a function is. :smile:
     
  17. Aug 12, 2011 #16

    I like Serena

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    I was looking for sin(3x+2pi) = sin(3(x+2pi/3))
    So if you add 2pi/3 to x, you'll get the same value for the sine function.
    This means the period is 2pi/3.


    Try again?
     
  18. Aug 12, 2011 #17

    PeterO

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    So when you said,

    If i take this case: sin(3x)=sin(3x+2pi), then 3x gets cancelled and i am left with 0=2pi. :(

    What how exactly were you going to cancel the 3x ?
     
  19. Aug 12, 2011 #18
    Because if sin x=sin y is given, can't we rewrite it as x=y?
     
  20. Aug 12, 2011 #19

    PeterO

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    NO. since sine is a periodic function, there is an infinite number of angles, which when put through the sine function give out the same value.

    For example, try the following

    sin 30
    sin 150
    sin 390
    sin 750

    There are lots of others that will give the same value.
     
  21. Aug 12, 2011 #20

    I like Serena

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    Nooooo.
    Since sine has a period of 2pi, this means its value is the same whenever you add 2pi to its argument.
    It's even worse, since sine takes on the same value twice in each period.

    The proper way to rewrite sin x=sin y is:
    x≡y (mod 2pi) or x≡pi-y (mod 2pi)
     
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