Problem 19 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\sin (2 x-\pi)$$

Step-by-Step Solution

Answer

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

Step 1: Finding the Amplitude

The amplitude of the sine function is the absolute value of the coefficient before the sine function. In the provided function \(y = \sin (2x - \pi)\), there is no coefficient before the sine function. Therefore, the amplitude of the function is 1.

Step 2: Finding the Period

The period of a sine function is given by \(2\pi / b\), where \(b\) is the coefficient of \(x\) in the sine function. Here, the coefficient of \(x\) is 2. So the period is \(2\pi /2 = \pi\) units.

Step 3: Finding the Phase Shift

The phase shift of a sine function can be found by setting the angle part of the sine equation equal to zero and solving for \(x\). In this case, the equation \(2x - \pi = 0\) can be solved to find that \(x = \pi /2\). Thus, the phase shift is \(\pi /2\) units.

Step 4: Graphing the Function

Now with the amplitude, period and phase shift, one can draw the sin graph. The amplitude tells us about the highest and lowest points of the wave, which are 1 and -1 respectively. The period \(\pi\) tells us the distance to complete one cycle, and the phase shift \(\pi /2\) tells us where the wave starts.