Question

consider the function ( y=cos (x) ). which change would increase the period by a factor of 3?
multiply ( cos (x) ) by ( \frac{1}{3} ).
multiply ( cos (x) ) by 3.
multiply ( x ) by ( \frac{1}{3} ).
multiply ( x ) by 3.

Explanation:

Step1: Recall the period formula for \(y = A\cos(Bx)\)

The period of the function \(y = A\cos(Bx)\) is \(T=\frac{2\pi}{|B|}\). For \(y = \cos(x)\), \(B = 1\) and \(T = 2\pi\).

Step2: Analyze each option

• Option 1: \(y=\frac{1}{3}\cos(x)\)

Here \(B = 1\), so the period \(T=\frac{2\pi}{1}=2\pi\). Multiplying \(\cos(x)\) by \(\frac{1}{3}\) changes the amplitude (vertical - stretch/shrink), not the period.

• Option 2: \(y = 3\cos(x)\)

Here \(B = 1\), so the period \(T=\frac{2\pi}{1}=2\pi\). Multiplying \(\cos(x)\) by \(3\) changes the amplitude (vertical - stretch/shrink), not the period.

• Option 3: \(y=\cos(\frac{1}{3}x)\)

Using the period formula \(T=\frac{2\pi}{|B|}\), with \(B=\frac{1}{3}\), we have \(T=\frac{2\pi}{\frac{1}{3}}=6\pi\). The original period of \(y = \cos(x)\) is \(2\pi\), and \(6\pi=3\times2\pi\).

• Option 4: \(y=\cos(3x)\)

Using the period formula \(T=\frac{2\pi}{|B|}\), with \(B = 3\), we have \(T=\frac{2\pi}{3}\). This is a decrease in the period.

Answer:

C. Multiply \(x\) by \(\frac{1}{3}\)