How do you find the amplitude, period, phase shift for $y = \cos 3x$?

Hint: This problem deals with finding the amplitude, period and the phase shift of the given function. The period of a periodic function is the interval between two matching points on the graph. In other words, it is the distance along the x-axis that the function has to travel before it starts to repeat its pattern. The basic sine and cosine functions have a period of $2\pi $, while tangent has a period of $\pi $.

Complete step-by-step answer:
Here let the given function $y = \cos 3x$ be $f(x)$.
$ \Rightarrow y = f(x)$
Any function is denoted in such a way, where $f(x) = A\cos \left( {Bx - C} \right) + D$
$ \Rightarrow y = A\cos \left( {Bx - C} \right) + D$
Here $A$ is the amplitude of the function.
Here P is the period of the function.
$ \Rightarrow P = \dfrac{{2\pi }}{B}$
Here the phase shift is given by:
$ \Rightarrow \dfrac{C}{B}$
Where D is the vertical shift.
The given function is $f(x) = \cos 3x$
Here $D = 0$, which means there is no vertical shift.
Here the amplitude of $f(x) = \cos 3x$ is:
$ \Rightarrow A = 1$
Here $B = 3$
The period of the function is given by:
$ \Rightarrow P = \dfrac{{2\pi }}{3}$
Phase shift is given by:
$ \Rightarrow \dfrac{C}{B}$
But here $C = 0$
Hence the phase shift would be 0.

Final Answer: The amplitude, period, phase shift for $y = \cos 3x$ are 1, $\dfrac{{2\pi }}{3}$ and 0 respectively.

Note:
Please note that the fundamental period of a function is the period of the function which are of the form, $f\left( {x + k} \right) = f\left( x \right)$ and $f\left( x \right) = f\left( {x + k} \right)$, then $k$ is called the period of the function and the function $f$ is called a periodic function. The period is the length of the smallest interval that contains exactly one copy of the repeating pattern. Any part of the graph that shows this pattern over one period is called a cycle.