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Finding two vlaues for trig function

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find at least two values of x in the interval [0,∏/2]
    for which f(x)= sin(x)+sin(2x)+sin(3x)=1


    3. The attempt at a solution

    now this is what my understanding is and this is what I have done


    sin(x)+sin(2x)+sin(3x)=1
    sin(x)+sin(2x)+sin(3x)-1

    Now by calculation f(0) = sin(0)+sin(2x0)+sin(3x0)-1 =-1

    and f(∏/2) = sin(∏/2)+sin(2x∏/2)+sin(3x∏/2)-1
    that then given me = 1+0+-1 -1 =-1

    but I know this is not right- as I don't get negative to positive solution to show that there is atleast two values of x for which f(x) is true!!!
    I think I have lost the plot somewhere!!!!!

    can someone look at this and tell me whats wrong and guide me to correct procedure
     
  2. jcsd
  3. Apr 11, 2012 #2

    SammyS

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    If the following is true,
    then
    Which says that f(x) - 1 = 0 .

    So the rest is wrong.
    Use the angle addition identity for sin(x+2x) which is sin(3x) .

    Similarly, for sin(x), you can think of x as being 2x - x .
     
  4. Apr 12, 2012 #3
    ok I am still lost
     
  5. Apr 12, 2012 #4
    Thank you I found similar example which cleared all my misunderstanding
     
    Last edited: Apr 12, 2012
  6. Apr 12, 2012 #5
    thia gives me
    2sin x + 2 sin x cos x - 2sin³ x + 2 cos² x sin x.
     
  7. Apr 12, 2012 #6

    SammyS

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    f(x) - 1 = sin(x)+sin(2x)+sin(3x) -1
    = sin(2x-x)+sin(2x)+sin(2x+x) -1

    = sin(2x)cos(x)-sin(x)cos(2x) + sin(2x) + sin(2x)cos(x)+sin(x)cos(2x) -1

    = ...​
     
  8. Jun 15, 2012 #7
    To expand further on this do you take say sin(2x) and break it down or do you start substituting values for x?
     
  9. Jun 15, 2012 #8

    SammyS

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    Hello Fizzman. Welcome to PF !

    Two of the terms cancel.

    Collect terms that have sin(2x), then see what happens if you expand sin(2x) .
     
  10. Jun 15, 2012 #9
    Thanks for the welcome.

    Ok so doing that I get

    2sin(2x)cos(x) +sin(x) -1

    Do I substitute for x or is further expansion needed?

    If I expand the sin(2x)

    2sin(2x)cos(x)+2sin(x)cos(x)-1
     
    Last edited: Jun 15, 2012
  11. Jun 15, 2012 #10

    SammyS

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    After looking at this problem for a while, it doesn't appear that it works out very nicely.

    I get f(x)=sin(2x)cos(x)-sin(x)cos(2x) + sin(2x) + sin(2x)cos(x)+sin(x)cos(2x) -1

         = sin(2x)(2cos(x) + 1) - 1

    You can expand sin(2x) out & get a term with cos2(x) --- which can be changed to 1 - sin2(x) --- but there still remains one term with cos(x) in it.

    That messes up the whole works!
     
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