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.
Best Answer
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 $\sin \left(\dfrac{\displaystyle \pi}{\displaystyle15}\right)$. $\\[5pt]$
$T_{\displaystyle6,0}\left(\dfrac{\displaystyle \pi}{\displaystyle 15}\right) = \dfrac{\displaystyle \pi}{\displaystyle 15} + \dfrac{\displaystyle -\left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^3}{\displaystyle 3!} + \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^5}{\displaystyle 5!}$ $\\[5pt]$
$\sin(12^{\circ}) \approx \dfrac{\displaystyle \pi}{\displaystyle 15} - \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^3}{\displaystyle 3!} + \dfrac{\displaystyle \left(\frac{\displaystyle \pi}{\displaystyle 15}\right)^5}{\displaystyle 5!}$ $\\[5pt]$