To solve this problem, we need to approximate $\cos(0.22)$ using a Taylor polynomial of degree 3 and then calculate the absolute error between this approximation and the exact value obtained using a calculator.

Step 1: Taylor Polynomial of Degree 3 for $\cos(x)$

The Taylor series expansion for $\cos(x)$ around $x = 0$ is given by:

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

For a Taylor polynomial of degree 3, we truncate this series to:

$P_3(x) = 1 - \frac{x^2}{2!}$

Step 2: Substitute $x = 0.22$

Now, substitute $x = 0.22$ into the Taylor polynomial:

$P_3(0.22) = 1 - \frac{(0.22)^2}{2}$

Let's compute this.

$P_3(0.22) = 1 - \frac{0.0484}{2} = 1 - 0.0242 = 0.9758$

So, the approximation of $\cos(0.22)$ using the Taylor polynomial $P_3(x)$ is approximately 0.975800.

Step 3: Calculate the Exact Value of $\cos(0.22)$

Using a calculator, we find:

$\cos(0.22) \approx 0.975897$

Step 4: Compute the Absolute Error

The absolute error is given by the difference between the exact value and the approximation:

$\text{Absolute Error} = |\cos(0.22) - P_3(0.22)| = |0.975897 - 0.9758| = 0.000097$

Final Answers:

• a. $P_3(0.22) = 0.975800$ (rounded to six decimal places)
• b. The absolute error is 0.000097.

Would you like more details, or do you have any questions on this problem?

Here are 5 related questions to expand your understanding:

1. How does increasing the degree of the Taylor polynomial affect the accuracy of the approximation?
2. What is the general formula for the nth-degree Taylor polynomial for $\cos(x)$?
3. How would you find the remainder term to estimate the error for higher-degree polynomials?
4. Can you approximate $\sin(0.22)$ using a third-degree Taylor polynomial?
5. How does the Taylor series for $\cos(x)$ behave as $x$ approaches larger values?

Tip: When approximating functions using Taylor series, always consider the interval where the series provides a good approximation and the role of higher-order terms in improving accuracy.