Taylor Series for sinx about pi/6

[mkwok]

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?

 

[Office_Shredder]

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

 

[mkwok]

I am not sure how that relates to Taylor's Series?