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Limit of trigonometry

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the limit


    2. Relevant equations
    lim θ→ ∏/6 (sinθ - 1/2)/(θ - ∏/6)

    3. The attempt at a solution

    well, if you substitute the values, it gives 0/0, but that means that is undefined and that's not the answer. according to my book the answer is √(3)/2 and I am not supposed to use L'Hopitals rule. this is we're I need help. Thanks
     
  2. jcsd
  3. Feb 17, 2014 #2

    Mark44

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    Use a substitution (u = θ - ##\pi##/6) to rewrite your limit as u → 0. Try to get it into the form ##\lim_{u \to 0}\frac{sin(u)}{u}##, which is a well-known limit.
     
  4. Feb 17, 2014 #3

    lurflurf

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    That is the derivative of sin(x) at x=∏/6. If you know the derivative of sine you are done. The derivative of sine can be expressed in terms of the derivative of sine when x=0. The derivative of sine at x=0 is usually an axiom or easily follows from the axioms.
     
  5. Feb 17, 2014 #4
    so basically the whole thing gives me 1? or the -1/2 stays and subtracts after i do the limit?
     
  6. Feb 17, 2014 #5

    Mark44

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    Make the substitution and see what you get.
     
  7. Feb 17, 2014 #6

    Mark44

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    I'm guessing that the aim of the problem is to get the poster to evaluate the limit, but not in the context that it's the derivative of anything.
     
  8. Feb 17, 2014 #7
    what i have right now is lim u->0 (sin(u) - (1/2))/u. after that I know that lim u->0 sin(u)/u= 1 then 1-(1/2). but it gives me 1/2 and that's not the correct answer. what am I doing wrong?
     
  9. Feb 17, 2014 #8
  10. Feb 17, 2014 #9

    SammyS

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    That's not what you should be getting.

    Show how you get that.
     
  11. Feb 17, 2014 #10
    that's how I did it... sin(u)/u= 1 so 1-(1/2)=1/2
     
  12. Feb 17, 2014 #11

    Mentallic

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    What's the expression you ended up with to conclude this? Show us each step you've taken, starting with the u substitution.
     
  13. Feb 17, 2014 #12

    SammyS

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    Mark 44 suggested that you use the substitution,
    ##\displaystyle u=\theta-(\pi/2) \ ## .​

    Of course, that gives
    ##\displaystyle \theta=u+(\pi/2) ##

    You need to use the angle addition formula for
    ##\displaystyle \sin\left(u+(\pi/2)\right)
    ##​
     
  14. Feb 20, 2014 #13
    but he said that u= theta - (pi/6) not pi/2
     
  15. Feb 20, 2014 #14
    according to what i have read, sin(u+(pi/2))= cos u but, can I do it with sin(u+(pi/6)) ?
     
  16. Feb 20, 2014 #15

    Mentallic

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    He made a mistake.

    No, you can't, but do you know how to expand sin(A+B)?
     
  17. Feb 20, 2014 #16
    yes sin(u+v)=sin(u)cos(v)+cos(u)sin(v) in which this case would be sin(u+(pi/2))= sin(u)cos(pi/2)+cos(u)sin(pi/2)
     
  18. Feb 20, 2014 #17

    SammyS

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    Yup, Mentallic is right.

    I can't read my own posts to proof-read them. :blushing:
     
  19. Feb 20, 2014 #18

    Mentallic

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    Ok, two problems. We're supposed to be making the substitution [itex]u = \theta - \pi/6[/itex] instead, and expressions like [itex]\cos(\pi/2)[/itex] can be simplified.
     
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