Use the table: x: 0, pi/2, pi, 3pi/2, 2pi

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

The values of the sine function at intervals of pi/2 have been recorded in the table. Identify the value of the phase shift (c) and write the functions equation.

A. f(x) = sin(x + pi/2)
B. f(x) = sin x + pi/2
C. f(x) = sin( x - pi/2)
D. f(x) = sin(x + pi)

1 answer

The values in the table represent the sine function evaluated at specific intervals:

  • \( F(0) = 1 \) which corresponds to \( \sin(0) = 0 \)
  • \( F(\frac{\pi}{2}) = 0 \) which corresponds to \( \sin(\frac{\pi}{2}) = 1 \)
  • \( F(\pi) = -1 \) which corresponds to \( \sin(\pi) = 0 \)
  • \( F(\frac{3\pi}{2}) = 0 \) which corresponds to \( \sin(\frac{3\pi}{2}) = -1 \)
  • \( F(2\pi) = 1 \) which corresponds to \( \sin(2\pi) = 0 \)

Given the values of the sine function, they suggest that it has been shifted in such a way that the sine function no longer starts at its usual position (0,1) at \(x=0\).

To analyze the phase shift, we see that the function values suggest that the peaks and troughs of the sine function have been horizontally shifted.

  1. The function starts at \( F(0) = 1 \) instead of \( 0 \) (the normal start of \( \sin(x) \)).
  2. The peak of the sine function at \( \frac{\pi}{2} \) aligns with \( y = 0 \), and zeros of the sine function happen at \( 0 \) and \( \pi \).

This corresponds to a downward shift of sine, implying a phase shift in the negative direction. Specifically, \( F(x) = \sin(x - \frac{\pi}{2}) \) would lead the sine curve to align where:

  • \( F(x) \) at \( x = 0 \) yields \( 1 \) (the shifted peak).
  • \( F = 0 \) at \( x = \frac{\pi}{2} \) (where sine is zero normally).

Upon reviewing the options,

  • Option A represents a left shift),
  • Option B suggests a vertical shift, not a phase,
  • Option C demonstrates a leftward phase shift,
  • Option D indicates a horizontal left continuous down,

Thus, from analysis, the equation of the sine function that matches this table is:

C: \( f(x) = \sin(x - \frac{\pi}{2}) \).