Question

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

Answer 1

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

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

Therefore

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

Answer 2

Answer:

Step-by-step explanation: