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Trigonometry exam problem

  1. May 29, 2008 #1
    I'm preparing for an entrance exam and i have this problem that is puzzling me.

    tan(5x) = tan(3x)

    I suppose i should start off by:

    sin(5x)/cos(5x) = sin(3x)/cos(3x)

    but i dont know what method to use.

    Should i break everything down to a single x? Please give me a hint how to start.
  2. jcsd
  3. May 29, 2008 #2


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    Sounds like a good idea to me.
  4. May 29, 2008 #3


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    explore this approach:

    tan(4x + x) = tan(4x - x)
  5. May 29, 2008 #4
    Use the identity 2sin(a)cos(b) = sin(a+b) + sin(a-b).

    *edit* Or, rather, derive an identity for tan(a) - tan(b) using algebra and knowledge of sin addition/subtraction formulas.
    Last edited: May 29, 2008
  6. May 29, 2008 #5
    Ok i solved it and i received something but i'm not sure if it is correct. May i post it here so you can check it if it's not too much trouble?
  7. May 29, 2008 #6
    What do you have? If my calculations are right, there should only be one value of x that satisfies the equation.
  8. May 29, 2008 #7
    I ended up with the quadratic equation:

    tan^2(4x)*tanx + tanx = 0

    tanx(tan^2(4x) + 1) = 0

    From here

    tanx = 0 // x = 0
    tan^2(4x) = -1 //Not possible
    Last edited: May 29, 2008
  9. May 29, 2008 #8
    Actually, sorry zero is not the only value, I made a logic error. Multiples of 2pi should be the answer. Why?
  10. May 29, 2008 #9
    Yes because the period of tan(x) function is 2pi.
  11. May 29, 2008 #10


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    Warning. The period of tan(x) is actually pi.
  12. May 29, 2008 #11
    So the solution is n*pi ?
  13. May 29, 2008 #12
    OK thank you.
  14. May 29, 2008 #13
    Hmm, sorry I meant multiples of pi, I solved it a different way and didn't check my answers thoroughly :-\.

    tan(a)-tan(b) = sin(a-b) / cos(a)*cos(b), we arrive at the equation

    sin(2x) / cos(3x)*cos(2x) = 0.

    sin(2x) = 0 when x = k(pi) and x = (2k+1)*pi/2 for all integers k. But tan(x) is undefined for the latter set of solutions, so x must be a multiple of pi.
  15. May 29, 2008 #14
    May i post another problem here or should i open a new thread?
  16. May 29, 2008 #15


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    Well, if tan(a)=tan(b) then a-b=n*pi for some integer n, because the period of tan is pi and tan assumes every value only once in it's period. Can you apply that to your problem?
  17. May 29, 2008 #16


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    It's up to you. But new threads attract more attention than old threads.
  18. May 29, 2008 #17
    Ok tnx for the tip and for all of your help about this problem :)
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