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 $\frac{\pi}{6}$?
 A: Normally, we have a Taylor Series for $\sin(x)$ centered around $x=0$:
$$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
or written more obviously, this is 
$$\sum_{n=0}^\infty (-1)^n\frac{(x-0)^{2n+1}}{(2n+1)!}$$
We know this is centered at zero because our computations of the derivatives and the $(x-0)$ term. But what does this mean? Well, it means that we approximate the curve $\sin(x)$ using higher and higher order derivatives at zero, so like the tangent line, the "tangent cubic", the "tangent quintic", and so on, as you can see here:

As you can see, at $x=0$, all the curves approximate $\sin(x)$ really well. The farther you get from $x=0$, the worse the approximation gets. Well, what if I want a value pretty far away? I would have to use many terms if I restricted myself to be centered at zero, but I can always center my approximation around a new point point, as you can see in this picture here:

This new Taylor series is the one you can compute using derivatives evaluated at $x=5\pi/6$ and the terms $(x-5\pi/6)^n$. All that we are doing is calculating the tangent line, the "tangent cubic", the "tangent quintic", and so on at the new point, allowing for more accurate calculations at and around this new point of $x=5\pi/6$ without having a gajillion terms in our sum.
Any questions just ask in the comments!
A: 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 $x-\frac{\pi }{6}$.
