Find the first four nonzero terms of the Taylor series for $\sin x$ centered at $\frac{\pi}6$ 
Find the first four nonzero terms of the series for $f(x)$ centered at $a$, using the definition of Taylor series.  $$f(x) = \sin(x),\quad a=\pi/6$$

I got this:
1st term: $1/2$
2nd: $\sqrt{3}/2$
3rd: $-1/2$
4th: $-\sqrt{3}/2$
but it seems I am very wrong, when I checked the answer. What am I doing wrong?
 A: One may recall that, for any sufficiently regular function $f$, by the Taylor series expansion near $x=a$, one has
$$
f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+(x-a)^3\varepsilon(x-a)
$$ with $\displaystyle \lim_{x \to a}\varepsilon(x-a)=0$. Here we take $f(x)=\sin x$, $a=\dfrac{\pi}6$, then classically
$$
f\left(\frac{\pi}6\right)=\frac12,\quad f'\left(\frac{\pi}6\right)=\frac{\sqrt{3}}2, \quad f''\left(\frac{\pi}6\right)=-\frac12, \quad f'''\left(\frac{\pi}6\right)=-\frac{\sqrt{3}}2,
$$ giving

$$
\sin x=\frac12+\frac{\sqrt{3}}2\left(x-\frac{\pi }{6}\right)-\frac14 \left(x-\frac{\pi }{6}\right)^2-\frac{\sqrt{3}}{12}\left(x-\frac{\pi }{6}\right)^3+\left(x-\frac{\pi }{6}\right)^3\varepsilon\left(x-\frac{\pi }{6}\right).
$$

A: Another way to do it could have been to set $x=y+\frac \pi 6$ and use
$$\sin(y+\frac \pi 6)=\frac{\sqrt{3}}{2}  \sin (y)+\frac{1}{2}\cos (y)$$ and, now, use Taylor series around $y=0$ $$\sin(y)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n+1}$$ $$\cos(y)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}y^{2n}$$ which make
$$\sin(y+\frac \pi 6)=\frac{\sqrt{3}}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}y^{2n+1}+\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}y^{2n}$$ Back to $x$ 
$$\sin(x)=\frac{\sqrt{3}}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x-\frac \pi 6)^{2n+1}+\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x-\frac \pi 6)^{2n}$$ 
