Question

X = 0, pi/2, pi, 3pi/2, 2pi

F(x) = 1, 0, -1, 0 ,1

The values of a siem function at intervals of pi/2 have been recorded. Identify the phase shift (c) and wrote the equation of the funcgion

Answer 1

The values you provided for the function \( F(x) \) at specified values of \( x \) suggest that the function is periodic and resembles a sine or cosine function. Given the values:

- At \( x = 0 \), \( F(0) = 1 \)
- At \( x = \frac{\pi}{2} \), \( F(\frac{\pi}{2}) = 0 \)
- At \( x = \pi \), \( F(\pi) = -1 \)
- At \( x = \frac{3\pi}{2} \), \( F(\frac{3\pi}{2}) = 0 \)
- At \( x = 2\pi \), \( F(2\pi) = 1 \)

These values match the sine function but with a phase shift. The sine function typically starts at \( 0 \) and goes through these values, but here you have the values of a cosine function shifted.

### Identify the Phase Shift

The standard cosine function, \( F(x) = \cos(x) \), starts at \( 1 \) when \( x = 0 \). In this case, it looks like we still have a cosine function but shifted to the right.

To determine the phase shift \( c \):

1. Standard cosine function: \( F(x) = A \cos(B(x-c)) + D \).
2. Here, you can see \( A = 1 \) (the amplitude), \( D = 0 \) (vertical shift), and the period of the sine or cosine function is \( 2\pi \).
3. The sine function \( F(x) = \sin(x) \) would go through \( 0 \) at \( \frac{\pi}{2} \). If we were using a cosine function, we could represent it as:

\[
F(x) = \cos(x - c)
\]

4. If we want \( F(0) = 1 \), we can hypothesize:

\[
x - c = 0 \Rightarrow c = 0
\]

Alternatively, we can equate each point:
\[
F(x) = \cos\left(x - \frac{\pi}{2}\right)
\]
Would give the sequence obtained in the function since the cosine wave will have the necessary amplitude and symmetry.

### Write the Function's Equation:

Given that it seems to be a cosine function with amplitude equal to 1 and a phase shift of:

\[
F(x) = \cos(x)
\]

If we want to write it in terms of shifted cosine function with two oscillations every \( 2\pi \):

\[
F(x) = \cos(x - c)
\]

Thus, replacing with the founded variables leads us to:

1. Original phase \( c = 0 \) which coherently provides:

\[
F(x) = \cos(x)
\]

2. The period is standard \( 2\pi \) without transformations.

Hence, the final equation can be denoted as:

\[
F(x) = \cos(x)
\]

This oscillates matching all values:

Thus, the phase shift \( c = 0 \) and the function:

\[
F(x) = \cos(x)
\]