
taylor polynomial
I need to find the thirddegree taylor polynomial at a for f(x) = sin9x and a = pi/54. I guess it's not clear to me what to do with the 9 in sin9x, does it end up in the final answer or only to calculate the coefficients? If someone could explain this polynomial intuitively (particularly what to do with the 9), that would be awesome.

$\displaystyle f(x) = \frac {f(a)}{0!}(xa)^{0} + \frac {f'(a)}{1!} (xa)^{1} + \frac {f''(a)}{2!}(xa)^{2} + \frac {f'''(a)}{3!}(xa)^{3} + ... $
$\displaystyle = f(a) + f'(a) (xa) + \frac {f''(a)}{2}(xa)^{2} + \frac {f'''(a)}{6}(xa)^{3} + ... $
$\displaystyle f(x) = \sin (9x) $
$\displaystyle f'(x) = 9 \cos(9x) $
$\displaystyle f''(x) = 81 \sin(9x) $
$\displaystyle f'''(x) = 729 \cos(9x) $
now find f(a), f'(a), f''(a), and f'''(a)

ok cool, I got that far, I just wanted to know what to enter into the (x  a) part of the polynomial. Do I replace x with 9x because of the 9 in sin(9x). In other words, would the final polynomial have (9x  a) to the nth power or just (x  a) to the nth power? For me, it's not clear what to do with the 9 and (most importantly) why.

just $\displaystyle \Big( x \frac {5 \pi}{4} \Big) $