Question

example 4 | a shifted sine curve
find the amplitude, period, and phase shift of ( y = 3 sin 2 left( x - \frac { pi } { 4 }
ight) ), and graph one com-
plete period.

Explanation:

Step1: Find the amplitude

The general form of a sine function is \(y = A\sin(B(x - C))+D\). For the function \(y = 3\sin(2(x-\frac{\pi}{4}))\), the amplitude \(|A|\). Here \(A = 3\), so the amplitude is \(|3|=3\).

Step2: Find the period

The formula for the period of a sine function \(y = A\sin(B(x - C))+D\) is \(T=\frac{2\pi}{|B|}\). Here \(B = 2\), so \(T=\frac{2\pi}{2}=\pi\).

Step3: Find the phase - shift

The formula for the phase - shift of a sine function \(y = A\sin(B(x - C))+D\) is \(C\). Here \(C=\frac{\pi}{4}\), so the phase - shift is \(\frac{\pi}{4}\) (to the right).

Step4: Graphing

We know that the key points of \(y=\sin x\) are \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2},- 1)\), \((2\pi,0)\).
For \(y = 3\sin(2(x-\frac{\pi}{4}))\), we make the substitution \(u = 2(x-\frac{\pi}{4})\) or \(x=\frac{u}{2}+\frac{\pi}{4}\).
When \(u = 0\): \(x=\frac{0}{2}+\frac{\pi}{4}=\frac{\pi}{4}\), \(y = 0\)
When \(u=\frac{\pi}{2}\): \(x=\frac{\pi/2}{2}+\frac{\pi}{4}=\frac{\pi}{2}\), \(y = 3\)
When \(u=\pi\): \(x=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\), \(y = 0\)
When \(u=\frac{3\pi}{2}\): \(x=\frac{3\pi/2}{2}+\frac{\pi}{4}=\pi\), \(y=-3\)
When \(u = 2\pi\): \(x=\frac{2\pi}{2}+\frac{\pi}{4}=\frac{5\pi}{4}\), \(y = 0\)

Answer:

Amplitude: \(3\), Period: \(\pi\), Phase - shift: \(\frac{\pi}{4}\) (to the right)