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I have a partial solution to the problem, but I am unsure of how to approximate a function to a desired number of decimal places, as opposed to approximating the function and error. Perhaps I am misunderstanding the question. If someone could clarify how to achieve this and the reasoning behind each step, I would appreciate it. Thank you.

The following calculations is my partial solution.

We first find $T_{\displaystyle6,0}(x)$ for $f(x) = \sin(x)$.$\\$

$f(x) = \sin (x)$ $\\$

$f'(x) = \cos (x)$ $\\$

$f''(x) = -\sin (x)$ $\\$

$f'''(x) = -\cos (x)$ $\\$

$f^4(x) = \sin (x)$ $\\$

$f^5(x) = \cos (x)$ $\\$

$f^6(x) = -\sin(x)$ $\\$

\begin{equation}T_{\displaystyle 6,a}(x) = \sin(a) + \frac{\displaystyle \cos(a)}{\displaystyle 1!}(x - a) + \frac{\displaystyle -\sin(a)}{\displaystyle 2!}(x - a)^2 + \frac{\displaystyle -\cos(a)}{\displaystyle 3!}(x - a)^3 + \\ \frac{\displaystyle \sin(a)}{\displaystyle 4!}(x - a)^4 + \frac{\displaystyle \cos(a)}{\displaystyle 5!}(x - a)^5 + \frac{\displaystyle -\sin(a)}{\displaystyle 6!}(x - a)^6 \\ \end{equation}

\begin{equation}\Rightarrow T_{\displaystyle6,0}(x) = \sin(0) + \frac{\displaystyle \cos(0)}{\displaystyle 1!}(x - 0) + \frac{\displaystyle -\sin(0)}{\displaystyle 2!}(x - 0)^2 + \frac{\displaystyle -\cos(0)}{\displaystyle 3!}(x - 0)^3 + \\ \frac{\displaystyle \sin(0)}{\displaystyle 4!}(x - 0)^4 + \frac{\displaystyle \cos(0)}{\displaystyle 5!}(x - 0)^5 + \frac{\displaystyle -\sin(0)}{\displaystyle 6!}(x - 0)^6 \\ \end{equation}

\begin{equation}\Rightarrow T_{\displaystyle6,0}(x) = \frac{\displaystyle 1}{\displaystyle 1!}(x) + \frac{\displaystyle -1}{\displaystyle 3!}(x)^3 + \frac{\displaystyle 1}{\displaystyle 5!}(x)^5 \\ \end{equation}

\begin{equation}\Rightarrow T_{\displaystyle6,0}(x) = x + \frac{\displaystyle -x^3}{\displaystyle 3!} + \frac{\displaystyle x^5}{\displaystyle 5!} \\ \end{equation}

We know that the function is equal to the Taylor Polynomial (our approximation of the function) and some remainder (the error in our approximation). In other words, $f(x) = \sin (x) = T_{\displaystyle6,0}(x) + R_{\displaystyle6,0}(x)$. $\\$

The function $f(x) = \sin (x)$ has continuous derivatives and is differentiable up to order $n + 1 (6 + 1)$ in the interval $x \in [-0.3, 0.3]$. Also, $a$ is an interior point of the interval $(a \in [-0.3, 0.3]$). $\\$

We now find the maximal possible error. $\\[5pt]$

$R_{\displaystyle6,0}(x) = \frac{\displaystyle -\cos(z)}{\displaystyle 7!}(x - 0)^7$ $\\[5pt]$

$|R_{\displaystyle6,0}(x)| = \frac{\displaystyle |-\cos(z)|}{\displaystyle 7!}|(x - 0)^7|$ $\\[5pt]$

$|R_{\displaystyle6,0}(x)| \le \frac{\displaystyle \max |-\cos(z)|}{\displaystyle 7!}\max|(x - 0)^7|$ $\\[5pt]$

$\Rightarrow |R_{\displaystyle6,0}(x)| \le \frac{\displaystyle 1}{\displaystyle 7!}|(0.3)^7|$ $\\[5pt]$

$\Rightarrow |R_{\displaystyle6,0}(x)| \le \frac{\displaystyle 1}{\displaystyle 5040}(0.0002187)$ $\\[5pt]$

$\Rightarrow |R_{\displaystyle6,0}(x)| \le 0.000000043$

We now approximate $\sin(12^{\circ})$ to $6$ decimal places.

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    $\begingroup$ The Taylor series is $f(x) = f(a) + f'(a) (x-a) + \frac {f''(a)(x-a)^2}{2!} \cdots$ If you are going to center your Taylor series at $0,$ then evaluate those trig functions at $0.$ It will make things look a bit cleaner. $\endgroup$ – Doug M Oct 6 '16 at 23:29
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    $\begingroup$ $12^\circ = \frac \pi{15}\approx 0.21$ radians, $f(x) = x - \frac {x^3}{3!} + \frac{x^5}{5!}$ and the next term will be less than $10^{-6}$ (It is actually on the order of $10^{-9}$) so you are inside your error bound. $\endgroup$ – Doug M Oct 7 '16 at 0:26
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    $\begingroup$ It the work you have already done. Set $a$ equal to $0. \sin (a) = 0, \cos(a) = 1$, and half of your terms go away. What you have done after that is correct. But you only need to consider the odd powered terms. $\endgroup$ – Doug M Oct 7 '16 at 0:33
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Thanks to Doug M's answers in the comments below the original post, I was able to understand and complete the solution.

We now approximate $\sin(12^{\circ})$ to $6$ decimal places. $\\$

$12^{\circ} = (12^{\circ}) \dfrac{\displaystyle \pi}{\displaystyle180^\circ}$ $\\[5pt]$

$\Rightarrow 12^{\circ} = \dfrac{\displaystyle \pi}{\displaystyle 15} $ radians $\\[5pt]$

We can see that $\dfrac{\displaystyle \pi}{\displaystyle 15} < 0.3$. In other words, it is within the interval shown above. Therefore, we can use the same Taylor Polynomial for $\sin (x)$, as shown above, to approximate $\sin \left(\dfrac{\displaystyle \pi}{\displaystyle15}\right)$. $\\[5pt]$

$T_{\displaystyle6,0}\left(\dfrac{\displaystyle \pi}{\displaystyle 15}\right) = \dfrac{\displaystyle \pi}{\displaystyle 15} + \dfrac{\displaystyle -\left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^3}{\displaystyle 3!} + \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^5}{\displaystyle 5!}$ $\\[5pt]$

$\sin(12^{\circ}) \approx \dfrac{\displaystyle \pi}{\displaystyle 15} - \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^3}{\displaystyle 3!} + \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^5}{\displaystyle 5!}$ $\\[5pt]$

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