Find the Taylor series of f(x)=sin(x)f(x) = \sin(x)f(x)=sin(x) at x=π/2x = \pi/2x=π/2.
∑n=0∞(x−π/2)2n(2n)!\sum_{n=0}^{\infty} \frac{(x-\pi/2)^{2n}}{(2n)!}∑n=0∞(2n)!(x−π/2)2n
∑n=0∞(−1)n(x−π/2)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi/2)^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n(x−π/2)2n+1
∑n=0∞(−1)n(x−π/2)nn!\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi/2)^n}{n!}∑n=0∞n!(−1)n(x−π/2)n
∑n=0∞(−1)n(x−π/2)2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi/2)^{2n}}{(2n)!}∑n=0∞(2n)!(−1)n(x−π/2)2n