$(10\mathrm{.}07HC)$

Consider the Maclaurin series: $g(x)=sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-$dots$+{\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$

Part A: Find the coefficient of the $4$th degree term in the Taylor polynomial for $f(x)=sin(3x)$ centered at $x=\frac{}{4}.(10$ points$)$

Part B: Use a $4$th degree Taylor polynomial for $cos(x)$ centered at $x=$ to estimate $g(3\mathrm{.}1).$ Explain why your answer is so close to $-1.(10$ points$)$

Part C: The series: ${\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$ has a partial sum $S_4=\frac{4241}{5040}$ when $x=1.$ What is an interval, $|S-S_4||R_4|$ for which the actual sum exists? Provide an exact answer and justify your conclusion. $(10$ points$)$