Question

What is the period of $f\left(x\right)=\frac{1}{2}\cos \left(3x\right)$ and then graph the function.

$\left(A\right)120{}^{\circ }\left(B\right)180{}^{\circ }\left(C\right)360{}^{\circ }\left(D\right)720°$

Short answer

The period of$f\left(x\right)=\frac{1}{2}\cos \left(3x\right)$ is $120°$. So, option (A) is correct.

Step-by-step solution

Step 1. Write down the given information.

The given function is $f\left(x\right)=\frac{1}{2}\cos \left(3x\right)$.

Step 2. Concept used.

A function of the form $y=a\sin \left(bx\right)\text{and}y=a\cos \left(bx\right)$ has amplitude of $\left|a\right|$ and period $\frac{360°}{\left|b\right|}\text{or}\frac{2\pi }{\left|b\right|}$.

Step 3. Evaluating period of the given function.

With the help of concept stated above, the period of the function is evaluated as:

The period of $f\left(x\right)=\frac{1}{2}\cos \left(3x\right)$ is $\frac{360°}{\left|3\right|}=120°$.

Step 4. Conclusion.

The period of $f\left(x\right)=\frac{1}{2}\cos \left(3x\right)$ is $120°$. So, option (A) is correct.