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Linear approximations derived from Taylor series

  1. Aug 10, 2009 #1
    1. The problem statement, all variables and given/known data

    So I have the problem questiona dn my teachers solution posted below.
    I understand:

    f(xo) = sin pi/6
    f '(xo) = cos pi/6

    but i dont know how he gets them into fraction form with the SQRT of 3, it looks like some pythagoras but i dont really know how he did it.


    2. Relevant equations

    Question 3(a) (sorry I named it 'answer' instead of 'question')

    answer.jpg

    3. The attempt at a solution

    solution.jpg

    :confused:
     
  2. jcsd
  3. Aug 10, 2009 #2

    tiny-tim

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    Hi _Greg_! :smile:

    (have a pi: π :wink:)
    I can't see the pictures yet, but π/6 = 30º, which is the little angle of half an equilateral triangle, so sin(π/6) (opp/hyp) = 1/2, and yes from Pythagoras you get cos(π/6) = √3/2. :smile:
     
  4. Aug 10, 2009 #3
    Thanks for the fast reply tiny tim.
    But how would you know its a 30* triangle, and how do you know cos pi/6 = sqrt3/2
    Like sin pi/6, is pi close enough to 3 to just call it 3/6?
    I don't really know much of the basics, I'm pretty much learning this for the deep end, looking for trends to understand
     
  5. Aug 10, 2009 #4

    tiny-tim

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    (what happened to that π i gave you? :confused:)
    Because π = 180º, so π/6 = 30º

    (and an equilateral triangle has all angles equal, and they add up to 180º, so each angle is 60º, and half that is 30º)
    No, just draw half an equilateral triangle …

    one side will be half the length of the other.
     
  6. Aug 10, 2009 #5
    aaaaaaah, I see, I thought that little n symbol was pi.
    And of course, it all adds up to 180*, high school maths is all coming back now :rolleyes:
    Many thanks :smile:
     
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