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.
- The function starts at \( F(0) = 1 \) instead of \( 0 \) (the normal start of \( \sin(x) \)).
- 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: \( f(x) = \sin(x - \frac{\pi}{2}) \).