Answer
Amplitude of the given function is $1$ , period is $2\pi $, and phase shift is $-\frac{\pi }{2}$.
Work Step by Step
For the standard function in the form $y=A\cos \left( Bx+C \right)$
$\begin{align}
& \text{Amplitude}=\left| A \right| \\
& \text{Period}=\frac{2\pi }{\left| B \right|}
\end{align}$
And,
$\text{Phase}\,\text{Shift}=-\frac{C}{B}$
Then, compare the given function $y=\cos \left( x+\frac{\pi }{2} \right)$ with the standard function as follows:
$A=1,\text{ }B=1,\text{ and }C=\frac{\pi }{2}$
So,
$\begin{align}
& \text{Amplitude=}\left| 1 \right| \\
& =1 \\
& \text{Period=}\frac{2\pi }{\left| 1 \right|} \\
& =2\pi
\end{align}$
And,
$\begin{align}
& \text{Phase}\,\text{Shift}=-\frac{\frac{\pi }{2}}{1} \\
& =-\frac{\pi }{2}
\end{align}$
And calculate quarter period
$\begin{align}
& \text{Quarter}\,\text{Period}=\frac{\text{Period}}{4} \\
& =\frac{2\pi }{4} \\
& =\frac{\pi }{2}
\end{align}$
Thus, the cycle begins at $x=-\frac{\pi }{2}$. Find the x-value and y-value at multiples of quarter periods.
Now, plot the obtained coordinates to get the graph of the function $y=\cos \left( x+\frac{\pi }{2} \right)$
