[Math] Find the first four nonzero terms of the Taylor series for $\sin x$ centered at $\frac{\pi}6$

Question

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?

Answer 1

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

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