taylor series sinx pi/6

Question

I am asked to compute the Taylor series of $\sin x$ about $\pi/6$. I know that you have to differentiate $f(x)=\sin x$ a few times and then compute $f^{(n)}(\pi/6)$ and find a pattern and then use the formula, but I am having a lot of trouble with finding a pattern that will fit.

Answer 1

First observe we have the trig identity \begin{align} \sin x = \sin \left(x-\frac{\pi}{6} + \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\sin \left(x-\frac{\pi}{6}\right)+\frac{1}{2}\cos \left(x-\frac{\pi}{6}\right). \end{align} Since the Maclaurin series for $\sin z$ and $\cos z$ are given by \begin{align} \sin z = \sum^\infty_{n=0} (-1)^n\frac{z^{2n+1}}{(2n+1)!} \ \ \text{ and }\ \cos z= \sum^\infty_{n=0} (-1)^n\frac{z^{2n}}{(2n)!} \end{align} then it follows \begin{align} \sin x = \frac{\sqrt{3}}{2}\sum^\infty_{n=0}(-1)^n\frac{(x-\pi/6)^{2n+1}}{(2n+1)!} + \frac{1}{2}\sum^\infty_{n=0}(-1)^n\frac{(x-\pi/6)^{2n}}{(2n)!}. \end{align}

Answer 2

Making the problem more general and using the same approach as Jacky Chong, if you want the Taylor expansion of $\sin(x)$ around $x=a$ use $$\sin(x)=\sin(x-a+a)=\sin(x-a)\cos(a)+\sin(a)\cos(x-a)$$ which would give $$\begin{align} \sin (x) = \cos(a)\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)!}(x-a)^{2n+1} + \sin(a)\sum^\infty_{n=0}\frac{(-1)^n}{(2n)!}(x-a)^{2n} \end{align}$$