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Homework Help: Basic calculus 1 question: What is the defining formula of sinΘ?

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Q) Derive the dereitvative of sinΘ using the definition of differentiation.

    I am in Calculus 3, and I used to know how to work this problem very well! :) I just don't remember lol - I just need a little help I guess! :)

    2. Relevant equations

    lim [f(x+h) - f(x))]/h

    3. The attempt at a solution

    f(x) = sinΘ

    f(x+h) = (sinΘ + h)

    Now I am STUCK!! lol

    THanks in Advance! :)
  2. jcsd
  3. Mar 2, 2010 #2


    User Avatar
    Homework Helper

    well now put that into the formula and check the numerator.

    You will have sin(θ+h)-sinθ. Do you know your sum to product formulas? These will help greatly here or you can expand sin(θ+h) as well.
  4. Mar 2, 2010 #3


    Staff: Mentor

    Might as well use x instead of theta, since x is easier to type.

    f(x + h) = sin(x + h)
    Use the identity for the sine of the sum of two angles.

    You will also need a limit:
    [tex]\lim_{h \to 0} \frac{sin(h)}{h}[/tex]
    Do you remember the value of this limit?
  5. Mar 2, 2010 #4
    so, I write:

    lim [sin(x+h) - sin(x))]/h

    So, I get:

    lim [sin(x) cos(h) + cos(x)sin(h) - sin(x))]/h

    But now I am stuck again lol - I understand that if I use [tex]\lim_{h \to 0} \frac{sin(h)}{h}[/tex]

    then my equation should become:

    lim [sin(x) cos(h) + cos(x)sin(h) ]/h , right?
  6. Mar 2, 2010 #5


    Staff: Mentor

    Your limit should be as h --> 0, not as x --> 0. Also, you are missing a term in the numerator. What happened to the -sin(x) that used to be there?

    There's another limit that you'll need as well:
    [tex]\lim_{h \to 0} \frac{cos(h) - 1}{h}[/tex]
  7. Mar 7, 2010 #6
    Took me a while for me to figure this out!! I spent time expanding on the ideas both of you gave me and finally understood what to do!!!

    THANKS to BOTH of you!!!!!!!!!! :)
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