Solution (a) The Maclaurin expansion for $y=\sin x$ was found in Example 3 (p. 616) of Section 8.9. $$y=\sin x=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}+\cdots=\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\dfrac{x^{2k+1} }{(2k+1) !}$$

Using the first four nonzero terms of the Maclaurin expansion, we can approximate $\sin x$ as $$\begin{equation*} \sin x\approx x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!} \tag{1} \end{equation*}$$

(b) The graphs of $y=\sin x$ and the approximation in (1) are given in Figure 29.

(c) Using (1), we get $$\sin 0.1\approx 0.1-\dfrac{0.1^{3}}{3!}+\dfrac{0.1^{5}}{5!}-\dfrac{0.1^{7}}{7!}\approx 0.0998$$

(d) Since the Maclaurin expansion for $y=\sin x$ at $x=0$ is an alternating series that satisfies the conditions of the Alternating Series Test, the error $E$ in using the first four terms as an approximation is less than or equal to the absolute value of the $5$th term at $x=0.1.$ That is, $$E\leq \left\vert \dfrac{0.1^{9}}{9!}\right\vert = 2.\, 756\times 10^{-15}$$