Find Taylor series for $\sin x$ at $\frac{\pi}{6}$ and estimate $\sin 38^\circ$ I want to see if I understand Taylor series correctly. Mind checking this work?
(a) So $f(x) = \sin{x}$ and $a = \frac{\pi}{6}$, $n = 4$ (I want to find $T_4(x)$ and I want to find the accuracy of the approximation when $x$ lies in the interval $[0,\frac{\pi}{3}]$.
So I write:
$$f'(x) = \cos{x},\quad
  f''(x) = -\sin{x},\quad
  f''(x) = -\cos{x},$$
$$
f^{(4)}(x) = \sin{x},\quad
f^{(5)}(x) = \cos{x}$$
and thus
  $$f'(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},\quad
f''(\frac{\pi}{6}) = \frac{-1}{2},\quad
f'''(\frac{\pi}{6}) = \frac{-\sqrt{3}}{2}$$
$$f^{(4)}(\frac{\pi}{6}) = \frac{1}{2},\quad
  f^{(5)}(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},$$
which gives
  $$T_4(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x-\frac{\pi}{6}) + \frac{\frac{-1}{2}(x-\frac{\pi}{6})^2}{2!} + \frac{\frac{-\sqrt{3}}{2}(x-\frac{\pi}{6})^3}{3!} + \frac{\frac{1}{2}(x-\frac{\pi}{6})^4}{4!}.$$
Question 1: Is the Taylor polynomial of the 4th degree above correct?
(b) Now I try to find the error in approximating $\sin\frac{\pi}{6}$ by the first 4 terms using Taylor's Inequality in my textbook:



First of all, $f^{(5)}(x) = \cos{x}$. For $x$ with $0 \leq x \leq \frac{\pi}{3}$, $\cos x$ is largest at $0$ so $f^{(5)}(x) \leq 1$ and $M = 1$.
So $$R_5(x) = \frac{1}{5!}(x-a)^5$$
I'm stuck here. 
Question 2: Can I say that $|x-a| < d$ where d = $\frac{\pi}{6}$?
(c) To estimate $\sin 38^{\circ}$, I write:
$$\sin 38^{\circ} = \sin(\frac{38}{180}\pi) = \sin(\frac{19}{90}\pi)$$
Question 3: Do I just plugin that value to the $T_4$ expression?
Question 4: More generally, what does it mean to be a Taylor series about $\frac{\pi}{6}$?

May be faster.
Let $x=y+\frac \pi 6$ making
$$\sin(x)=\sin \left(y+\frac{\pi }{6}\right)=\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)$$ for which the derivatives at $y=0$ are very simple.
As a result
$$\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)= \sum_{n=0}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi  n}{2}\right)+\cos \left(\frac{\pi 
   n}{2}\right)}{2 (n!)}\, y^n$$
Replace $y$ by $x-\frac{\pi }{6}$.
