Question

expand sinx in powers of (x-1/2x) by using Taylor's series​

Answer 1

Answer:

Expand sin x in powers of (x - π/2) by using Taylor's series.

Answer

Answer: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$sin^{(n)}(\frac{\pi}{2}) = \pm cos(\frac{\pi}{2}) = 0$ otherwiseTherefore$sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}$

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

Answer ·

The expansion of sin(x) in powers of (x-\frac{\pi}{2}) using