Question

expand sin x in powers of x π 2 by taylor's series

Answer 1

The expansion of sin(x) in powers of (x-\frac{\pi}{2}) using Taylor's series is

sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}

Step-by-step explanation:

By Taylor's Theorem,

sin(x) = sin(\frac{\pi}{2})+sin'(\frac{\pi}{2})(x-\frac{\pi}{2})+ \dots + \frac{sin^{(k)}(\frac{\pi}{2})}{k!}(x-\frac{\pi}{2})^k

But

sin^{(n)}(\frac{\pi}{2}) = sin(\frac{\pi}{2}) = 1 if n = 4k

sin^{(n)}(\frac{\pi}{2}) = -sin(\frac{\pi}{2}) = -1 if n = 4k+2

and

sin^{(n)}(\frac{\pi}{2}) = \pm cos(\frac{\pi}{2}) = 0 otherwise

Therefore