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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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