Question

expand sinx in power of x-pi/2 and find sin91

Answer 1

$\sf{ \blue{✿Hope \: this \: helps! }}$

Answer 2

The expansion of sin(x)sin(x) in powers of$(x-\frac{\pi}{2})(x−2π)$using Taylor's series is $\\ sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}sin(x) \\ =∑k∈N(4k)!(x−2π)4k−(4k+2)!(x−2π)4k+2$By Taylor's Theorem, $\\ sin(x) \\ = sin(\frac{\pi}{2})+sin'(\frac{\pi}{2})(x-\frac{\pi}{2})+ \dots + \frac{sin^{(k)}(\frac{\pi}{2})}{k!}(x-\frac{\pi}{2})^ksin(x) \\But \\ sin^{(n)}(\frac{\pi}{2}) = sin(\frac{\pi}{2}) \\ sin^{(n)}(\frac{\pi}{2}) = -sin(\frac{\pi}{2}) = -1sin(n)(2π) \\ and \\ sin^{(n)}(\frac{\pi}{2}) = \pm cos(\frac{\pi}{2}) = 0sin(n)(2π)=±cos(2π)=0  \: otherwise \\Therefore \\ sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}sin(x)$