Chapter 4.3, Problem 80E

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.

  $\begin{array}{cccc}\theta & \cos \theta & 1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}& \cos \theta -\left(1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}\right)\\ -0.3& -\mathrm{0.955..}& & \\ -0.2& -\mathrm{0.980..}& & \\ -0.1& -\mathrm{0.995..}& & \\ 0& 1& & \\ 0.1& \mathrm{0.995..}& & \\ 0.2& \mathrm{0.980..}& & \\ 0.3& \mathrm{0.955..}& & \end{array}$

The table shows that the Taylor polynomials is a good approximation for $\cos \theta$ and get better as $\theta$ approaches to $0$ and gives an accurate prediction for $\theta =0$ .

Given information:

The given table is:

  $\begin{array}{cccc}\theta & \cos \theta & 1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}& \cos \theta -\left(1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}\right)\\ -0.3& -\mathrm{0.955..}& & \\ -0.2& -\mathrm{0.980..}& & \\ -0.1& -\mathrm{0.995..}& & \\ 0& 1& & \\ 0.1& \mathrm{0.995..}& & \\ 0.2& \mathrm{0.980..}& & \\ 0.3& \mathrm{0.955..}& & \end{array}$

Explanation:

Consider the given information:

By the Taylor expansion.

Cosine function can be expanded as follows:

Complete the third column of the table as:

  $\cos \theta =1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}-\mathrm{....}$

For the value of $\theta =-0.3$ the value is as follows.

  $\begin{array}{c}\cos \left(-0.3\right)=1-\frac{{\left(-0.3\right)}^2}{2}+\frac{{\left(-0.3\right)}^4}{24}\\ \mathrm{0.955...}=0.9553375\end{array}$

This is true.

Similarly put the other values of $\theta$ .

The third column of the table filled is as follows:

  $\begin{array}{cccc}\theta & \cos \theta & 1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}& \cos \theta -\left(1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}\right)\\ -0.3& -\mathrm{0.955..}& & \\ -0.2& -\mathrm{0.980..}& & \\ -0.1& -\mathrm{0.995..}& & \\ 0& 1& & \\ 0.1& \mathrm{0.995..}& & \\ 0.2& \mathrm{0.980..}& & \\ 0.3& \mathrm{0.955..}& & \end{array}$

Further calculate the last column of the table.

  $\begin{array}{cccc}\theta & \cos \theta & 1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}& \cos \theta -\left(1-\frac{{\theta }^2}{2}+\frac{{\theta }^4}{24}\right)\\ -0.3& -\mathrm{0.955..}& 0.9553375& -0.00000101\\ -0.2& -\mathrm{0.980..}& 0.980067& -0.000067\\ -0.1& -\mathrm{0.995..}& 0.9950042& -0.0000042\\ 0& 1& 1& 0\\ 0.1& \mathrm{0.995..}& 0.9950042& -0.0000042\\ 0.2& \mathrm{0.980..}& 0.980067& -0.000067\\ 0.3& \mathrm{0.955..}& 0.9553375& -0.00000101\end{array}$

The table shows that the Taylor polynomials is a good approximation for $\cos \theta$ and get better as $\theta$ approaches to $0$ and gives an accurate prediction for $\theta =0$ .