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The problem involves using the Newton-Raphson method to find the roots of the function \( f(x) = \cos(x) \). Let's go through each part of the problem.

### Part (a): Newton-Raphson Formula
The Newton-Raphson method is given by the formula:

\[
p_{n+1} = p_n - \frac{f(p_n)}{f'(p_n)}
\]

For the function \( f(x) = \cos(x) \), we first need to find the derivative \( f'(x) \):

\[
f'(x) = -\sin(x)
\]

Substituting \( f(x) \) and \( f'(x) \) into the Newton-Raphson formula:

\[
p_{n+1} = p_n - \frac{\cos(p_n)}{-\sin(p_n)} = p_n + \frac{\cos(p_n)}{\sin(p_n)} = p_n + \cot(p_n)
\]

Thus, the Newton-Raphson formula for this function is:

\[
p_{n+1} = p_n + \cot(p_n)
\]

### Part (b): Finding the Root \( p = 3\pi/2 \)
To determine if we can use \( p_0 = 1 \) as an initial guess, we need to consider the behavior of the Newton-Raphson method. The root \( p = \frac{3\pi}{2} \) corresponds to the point where \( \cos(x) = 0 \), i.e., \( x = \frac{3\pi}{2} \).

Let's evaluate the derivative at \( p_0 = 1 \):

\[
f(1) = \cos(1) \approx 0.5403
\]
\[
f'(1) = -\sin(1) \approx -0.8415
\]

The next approximation \( p_1 \) would be:

\[
p_1 = 1 - \frac{0.5403}{-0.8415} \approx 1 + 0.6421 \approx 1.6421
\]

While this is a step closer, it's not close enough to \( \frac{3\pi}{2} \approx 4.712 \), and the Newton-Raphson method may converge slowly or to the wrong root. \( p_0 = 1 \) is not a good initial guess to find the root \( \frac{3\pi}{2} \).

### Part (c): Finding the Root \( p = 3\pi/2 \)
Now, let's check if we can use \( p_0 = 5 \) as the initial guess. Since \( 5 \) is closer to \( \frac{3\pi}{2} \), let's evaluate the derivative at \( p_0 = 5 \):

\[
f(5) = \cos(5) \approx 0.2837
\]
\[
f'(5) = -\sin(5) \approx 0.9589
\]

The next approximation \( p_1 \) would be:

\[
p_1 = 5 - \frac{0.2837}{0.9589} \approx 5 - 0.2958 \approx 4.7042
\]

This value is much closer to \( \frac{3\pi}{2} \approx 4.712 \). Therefore, \( p_0 = 5 \) is a better initial guess, and it is likely to converge to the root \( p = \frac{3\pi}{2} \).

### Summary
1. The Newton-Raphson formula for \( f(x) = \cos(x) \) is \( p_{n+1} = p_n + \cot(p_n) \).
2. \( p_0 = 1 \) is not a suitable initial guess for finding the root \( p = \frac{3\pi}{2} \).
3. \( p_0 = 5 \) is a suitable initial guess for finding the root \( p = \frac{3\pi}{2} \).

Do you want further details or have any questions? 

Here are 5 related questions:
1. How does the choice of initial guess affect the convergence of the Newton-Raphson method?
2. What happens if the derivative \( f'(x) \) is close to zero at the initial guess in Newton-Raphson?
3. Can the Newton-Raphson method fail to converge? Under what conditions?
4. How can the Newton-Raphson method be modified to handle functions with multiple roots?
5. How does the convergence speed of the Newton-Raphson method compare to other root-finding methods?

**Tip:** Always choose an initial guess close to the actual root to increase the likelihood of convergence in the Newton-Raphson method.

Consider the function f(x) = cos(x). a) Find the Newton-Raphson formula. b) To find the root p = 3π/2, can we use p₀ = 1? If so, why? c) To find the root p = 3π/2, can we use p₀ = 5? If so, why?

Explore how to apply the Newton-Raphson method to find the roots of cos(x). Learn how initial guesses affect convergence, specifically for p = 3π/2, with an analysis of p₀ = 1 and p₀ = 5. Ideal for university students studying calculus.

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