Problem 17 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\sin (x-\pi)$$

Step-by-Step Solution

Answer

The amplitude of the function \(y=\sin (x-\pi)\) is 1, the period is \(2\pi\), and the phase shift is \(\pi\) in the positive direction.

Step 1: Determining the Amplitude

The coefficient of sine function is 1 in the given function \(y=\sin (x-\pi)\). Thus, the amplitude is 1.

Step 2: Determining the Period

There's no coefficient in front of \(x\) in the given function, hence, the period of the given function remains the same as the sine function, which is \(2\pi\). Therefore, the period is \(2\pi\).

Step 3: Determining the Phase Shift

The function is shifted \(\pi\) units to the right as there's a subtraction in the argument of the function. Hence, the phase shift is \(\pi\).

Step 4: Graphing the Function

Begin marking the phase shift on the x-axis, at \(x=\pi\). Then, moving in both directions, make a mark for each additional period of \(2\pi\). Since the amplitude is 1, the maximum value of the function will be 1 and minimum will be -1. Plot the points and join them to complete one period of the function. Considering the nature of sine function, it starts from zero at the beginning of its period; here it will start from zero at \(x=\pi\) due to phase shift.