Approximate \(y=\sin x\) by using the first four nonzero terms of its Maclaurin expansion.

Graph \(y=\sin x\) along with the approximation found in (a).

Use (a) to approximate \(\sin 0.1.\)

What is the error in using this approximation?

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, \[ E\leq \left\vert \dfrac{0.1^{9}}{9!}\right\vert = 2.\, 756\times 10^{-15} \]

Problem 1.