Taylor Series for sinx about pi/6

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Homework Statement


Determine the Taylor Series for f(x)=sinx about the center point c=pi/6

Homework Equations


pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3! + ...

The Attempt at a Solution



f(pi/6) = 1/2
f'(pi/6) = \sqrt{3}/2
f''(pi/6) = -1/2
f'''(pi/6) = -\sqrt{3}/2
f(4)(pi/6) = 1/2 and the coefficients repeats from hereWhat I am stuck on is.. how do I make it so that (-1) will only appear on the 3rd and 4th term... and how do I make it so that \sqrt{3} will only appear on the odd terms?
 
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If w=x-pi/6 then sinx=sin(w+pi/6)=sin(w)cos(pi/6)+sin(pi/6)cos(w). Try expanding this and see what you get
 
I am not sure how that relates to Taylor's Series?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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