Chapter 4.3, Problem 79E

Solution

To fill: The values in the table. It is given that radian measure allows the trigonometric functions to be approximated by simple polynomial functions. Sine and cosine are approximated by Taylor polynomials. Complete each table showing a Taylor polynomial in the third column and identify the relative error.

  $\begin{array}{cccc}\theta & \sin \theta & \theta -\frac{{\theta }^3}{6}& \sin \theta -\left(\theta -\frac{{\theta }^3}{6}\right)\\ -0.3& -\mathrm{0.295..}& & \\ -0.2& -\mathrm{0.198..}& & \\ -0.1& -\mathrm{0.099..}& & \\ 0& 0& & \\ 0.1& \mathrm{0.099..}& & \\ 0.2& \mathrm{0.198..}& & \\ 0.3& \mathrm{0.295..}& & \end{array}$

Relative error is less than 1% for $\left|\theta \right|<1$ and approximation is better for $\theta$ is close to $0$ .

Given information:

The given table is:

  $\begin{array}{cccc}\theta & \sin \theta & \theta -\frac{{\theta }^3}{6}& \sin \theta -\left(\theta -\frac{{\theta }^3}{6}\right)\\ -0.3& -\mathrm{0.295..}& & \\ -0.2& -\mathrm{0.198..}& & \\ -0.1& -\mathrm{0.099..}& & \\ 0& 0& & \\ 0.1& \mathrm{0.099..}& & \\ 0.2& \mathrm{0.198..}& & \\ 0.3& \mathrm{0.295..}& & \end{array}$

Explanation:

Consider the given information.

Complete the table as shown below:

  $\begin{array}{cccc}\theta & \sin \theta & \theta -\frac{{\theta }^3}{6}& \sin \theta -\left(\theta -\frac{{\theta }^3}{6}\right)\\ -0.3& -0.295& -0.2955& -0.0000202\\ -0.2& -0.198& -0.198667& -2.664\times {10}^{-6}\\ -0.1& -0.099& -0.099833& -8.331\times {10}^{-8}\\ 0& 0& 0& 0\\ 0.1& 0.099& 0.099833& 8.331\times {10}^{-8}\\ 0.2& 0.198& 0.198667& 2.664\times {10}^{-6}\\ 0.3& 0.295& 0.2955& 0.0000202\end{array}$

See following figure to understand approximation by Taylor polynomials.

  

Observe the graph of $\sin \theta$ and Taylor polynomials $\theta -\frac{{\theta }^3}{6}$ it shows that it is a good approximation for $\sin \theta$ .

Relative error is less than 1% for $\left|\theta \right|<1$ and approximation is better for $\theta$ is close to $0$ .