Problem 18 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The amplitude of the function \(y=\sin \left(x-\frac{\pi}{2}\right)\) is 1, the period is \(2\pi\), and the phase shift is \(\frac{\pi}{2}\) units to the right. See the step-by-step solution for an illustration of how to graph one period of the function.

Step 1: Identify the Amplitude

The standard form of the sine function is \(y=A\sin(Bx - C)\), where |A| is the amplitude. In the given function \(y=\sin \left(x-\frac{\pi}{2}\right)\), there is no coefficient in front of the sine function, which implies the amplitude (|A|) is 1.

Step 2: Identify the Period

The period of the standard sine function \(y=A\sin(Bx - C)\) is \(\frac{2\pi}{|B|}\). In the given function \(y=\sin \left(x-\frac{\pi}{2}\right)\), the coefficient of \(x\) inside the sine function is \(1\), which means the period is \(\frac{2\pi}{1} = 2\pi\).

Step 3: Identify the Phase Shift

The phase shift of the standard sine function \(y=A\sin(Bx - C)\) is \(\frac{C}{B}\). In the function \(y=\sin \left(x-\frac{\pi}{2}\right)\), since the term inside the parentheses is \(x - \frac{\pi}{2}\), the phase shift is \(\frac{\pi}{2}\) units to the right.

Step 4: Graphing the function

To graph the function, plot the critical points of one period, which are a full cycle of a sine function. These include the midline, maximum, minimum, and zero-crossings. The sine function starts from the midline, rises to its maximum at \(\frac{\pi}{2}\), returns to the midline at \(\pi\), falls to its minimum at \(\frac{3\pi}{2}\) and ends the period at \(2\pi\) at the midline. The graph is shifted \(\frac{\pi}{2}\) units to the right due to the phase shift.