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Proving slope m of a secant connecting two points of the sine curve

  1. Mar 13, 2007 #1
    Proving slope "m" of a secant connecting two points of the sine curve

    1. The problem statement, all variables and given/known data

    Write and expression for the slope m of the secant connecting the points Po(Xo,Yo) and P(X,Y) of the sine curve. Use the appropriate trigonometric identity to show that m= sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

    2. Relevant equations

    Could somebody give a hint how to do the second proof? m= sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

    Basically what i need to prove is that:
    (Sin(Xo)-Sin(X)) / Xo-X = sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

    3. The attempt at a solution

    I could manage to do the first part of the exercise, as far as now I got a formula which is the following:

    m= (Yo-Y)/(Xo-x) = (Sin(Xo)-Sin(X)) / Xo-X
    and the expression (Sin(Xo)-Sin(X)) / Xo-X is equal to m
     
  2. jcsd
  3. Mar 13, 2007 #2

    Dick

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    Can you find a formula for sin(a)-sin(b)? It's pretty standard.
     
  4. Mar 13, 2007 #3
    Do you mean sin(A-B) = sinAcosB - cosAsinB ?
     
  5. Mar 13, 2007 #4

    Dick

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    That's related, but I really do mean sin(a)-sin(b). Look at an extensive table of trig formulae.
     
    Last edited: Mar 13, 2007
  6. Mar 13, 2007 #5

    Dick

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    Or if you want to prove it from scratch, add or subtract your formulas for sin(a+b) and sin(a-b).
     
  7. Mar 13, 2007 #6
    oohhh, i found the formula in my exercise book, it was really that standard

    thank you for your help, now im doing the (b) part of the exercise about their limits
    May I ask for your help if i get stuck?
     
  8. Mar 13, 2007 #7
    Im doing the (b) part of the exercise, which ask about what do they approach as X approaches Xo (their limits)

    1)sin((X-Xo)/2)/((X-Xo)/2)
    2)cos ((X+Xo)/2)
    3)m(=sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

    Note: Xo an X are points on the cos,sin curve which is not bigger than pi/2

    I was able to solve the first one, I got as X approaches Xo it will approach 1.(I think I did right, but please tell me if it's wrong)
    I got stuck in the second one, could you guys give me some hint to solve the 2.?
    In the thrid I guess all I need to do is to multiply the two limits (1. and 2.) and that's it.

    Any helps appreciated
     
    Last edited: Mar 13, 2007
  9. Mar 13, 2007 #8

    Dick

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    The second one is MUCH easier than the first one. X -> X0. No fancy limits to take really. The answer is a function of X0 - it's not a constant like the first one, if that is what is throwing you off.
     
  10. Mar 13, 2007 #9
    I don't really understand what you mean, could you please explain it in details?
     
  11. Mar 13, 2007 #10

    Dick

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    What's the limit of cos((x+2)/2) as x->2?
     
  12. Mar 13, 2007 #11
    isnt it approaches 1?
     
  13. Mar 13, 2007 #12

    Dick

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    Nope. As x->2, cos((x+2)/2)->cos((2+2)/2)=cos(4/2)=cos(2). Why 1?
     
  14. Mar 13, 2007 #13
    But dont you need to take the cosine of two?
     
  15. Mar 13, 2007 #14

    Dick

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    I could, but it isn't 1. cos(2*pi)=1, not cos(2). Do you agree the answer is cos(2)?
     
  16. Mar 13, 2007 #15
    I see what you mean, that (2) is in radian and not degrees, thus the answer to the original question is (cos(Xo))/2 right?
     
  17. Mar 13, 2007 #16

    Dick

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    It's cos((X0+X0)/2), but that doesn't simplify to what you wrote.
     
  18. Mar 13, 2007 #17
    Im getting a lost a little bit, is it actually approaches zero?
     
  19. Mar 13, 2007 #18

    Dick

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    So am I. Explain how you got cos(X0)/2???
     
  20. Mar 13, 2007 #19
    oooh if x-->Xo therefore is it (cos(2*Xo))/2?
     
  21. Mar 13, 2007 #20

    Dick

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    You are still being sloppy. The '/2' is INSIDE the cosine - not outside.
     
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