math

lim delta x to 0 of

sin[(pi/6)+delta x]-(1/2)
all divided by delta x

hope i made that clear.

hint given: sin(theta+gamma)=sin theta cos gamma+cos theta sin gamma

sorry not sure where to start on this one

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  1. first of all, to make things easier to type, let's say
    h = delta x

    so your question becomes

    lim [ sin(π/6 + h) - sin π/6 ]/h , as h ---> 0

    This looks like you want the derivative of
    sin x, when x = π/6

    if f(x) = sinx
    f'(x) = cosx
    f'(π/6) = cos π/6 = √3/2

    HOWEVER, it looks they actually want you to evaluate this by First Principles.
    Using the hint they gave you, this will be a messy messy calculation.
    Here is a page where they show those steps, they use d instead of h, and you will have to replace x with π/6
    http://www.math.com/tables/derivatives/more/trig.htm

    A more tradional method is to use
    sin A − sin B = 2 sin ½ (A − B) cos ½ (A + B)
    as seen in this video
    http://www.youtube.com/watch?v=T0HPGrUXNzY
    (hope you can understand the Indian accent, also close the ad at the beginning)

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