To apply Newton's Method, the formula for each iteration is:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

We are given the function $f(x) = \cos(x)$, and we need its derivative $f'(x)$.

Since $f(x) = \cos(x)$, we know that:

$f'(x) = -\sin(x)$

We are starting with the initial guess $x_1 = 1.9$.

Let's calculate two iterations of Newton's Method, rounding values to four decimal places. I'll create a table for this.

First, we summarize the required columns:

• $x_n$ (current estimate)
• $f(x_n)$ (function value at $x_n$)
• $f'(x_n)$ (derivative value at $x_n$)
• $\frac{f(x_n)}{f'(x_n)}$
• $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ (next estimate)

Initial Guess:

• $x_1 = 1.9$

Let’s calculate the values:

Iteration 1 (n = 1):

• $f(x_1) = \cos(1.9)$
• $f'(x_1) = -\sin(1.9)$
• Calculate $\frac{f(x_1)}{f'(x_1)}$
• Update $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$

Iteration 2 (n = 2):

• $f(x_2) = \cos(x_2)$
• $f'(x_2) = -\sin(x_2)$
• Calculate $\frac{f(x_2)}{f'(x_2)}$
• Update $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}$

Let’s compute these values now.Here is the table with two iterations of Newton's Method for approximating a zero of $f(x) = \cos(x)$ starting with $x_1 = 1.9$:

\hline n & x_n & f(x_n) & f'(x_n) & \frac{f(x_n)}{f'(x_n)} & x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \\ \hline 1 & 1.9000 & -0.3233 & -0.9463 & 0.3416 & 1.5584 \\ 2 & 1.5584 & 0.0124 & -0.9999 & -0.0124 & 1.5708 \\ \hline \end{array}$$ - After the first iteration, $$x_2 \approx 1.5584$$. - After the second iteration, $$x_3 \approx 1.5708$$. This process is converging to a root of $$\cos(x) = 0$$, which is close to $$x = \frac{\pi}{2} \approx 1.5708$$. --- Would you like more details, or do you have any questions? ### Follow-up Questions: 1. How does Newton's Method work with other types of functions? 2. What happens if we start with a different initial guess? 3. Can Newton's Method fail to converge? 4. Why is the derivative necessary in Newton's Method? 5. What is the geometric interpretation of Newton's Method? ### Tip: For faster convergence, the initial guess should be close to the actual root.