Find a formula for the nth term of the Taylor series of f(x), centered at c, by finding the coefficients of the first few powers of x and looking for a pattern.
\[p_n = \dfrac{f^{n}(c)}{n!}\]
\[f(π/2) = cos(π/2) = 0, p_0 = 0\]
\[f'(x) = -sinx\]
\[f'(π/2) = -1, p_1 = -1\]
\[f''(x) = -cosx\]
\[f''(π/2) = 0, p_2 = 0\]
\[f'''(x) = sinx\]
\[f'''(π/2) = 1, p_3 = \dfrac{1}{3!}\]
\[f''''(x) = cosx\]
\[f''''(x) = 0, p_4 = 0\]
Keeps on repeating past this point to achieve the pattern of Taylor coefficients 0, -1, 0, 1, 0, -1, 0, 1,…. Thus the taylor series is:
\[f(x) = \sum^{\infty}_{n=0}p_n(x-c)^n\]
\[T(x) = - (x-\dfrac{π}{2}) + \dfrac{1}{6} (x -\dfrac{π}{2})^3 - \dfrac{1}{5!}(x - \dfrac{π}{2})^5 + ...\]
\[= \sum^{\infty}_{k=0}(-1)^{k+1}\dfrac{1}{(2k+1)}(x-\dfrac{π}{2})^{2k+1}\]