Question

How do you graph and list the amplitude, period, phase shift for $y=-\sin \left(x-\pi \right)$ ?

Answer 1

Hint: First, using the suitable trigonometric identities, simplify the given equation and try to get a simplest form of the equation so that we can understand it better. Then find the maximum value of the function.

Complete step by step solution:
The given equation is $y=-\sin \left(x-\pi \right)---------\left(1\right)$
This is a trigonometric equation. All trigonometric functions are periodic. This means that the function repeats itself after a regular interval on the Cartesian plane.
The trigonometric function $\sin x$ has a period of $2\pi$ radians. This means that the values of the function $\sin x$ repeat after every interval of $2\pi$ radians.
This helps in graphing the curve of a trigonometric function. We can graph the function for an interval of $2\pi$ radians and then just replicate the function for every such successive interval.
Let us simplify equation $1$ by using the identity $-\sin \theta =\sin \left(-\theta \right)$ .
Then,
 $\Rightarrow y=-\sin \left(x-\pi \right)$
 $\Rightarrow y=\sin \left[-\left(x-\pi \right)\right]$
 $\Rightarrow y=\sin \left(\pi -x\right)$
Now, we shall use the identity $\sin \left(\pi -x\right)=\sin x$
So, we get, $y=-\sin \left(x-\pi \right)=\sin x$
This means that the graph of equation $\left(1\right)$ is the same as the graph of trigonometric function $\sin x$ .
So, we get the graph of $y=-\sin \left(x-\pi \right)$ as

So, we now know that $y=-\sin \left(x-\pi \right)=\sin x$
Hence, the maximum value of the function $y=-\sin \left(x-\pi \right)$ is $1$ .
Therefore, the amplitude of the function $y=-\sin \left(x-\pi \right)$ is $1$ .
Period of the function $y=-\sin \left(x-\pi \right)$ is $2\pi$ radians.
Phase shift of the graph is zero.

Note: If we have an equation $A\sin \left(kx-\varphi \right)$ , then A is the amplitude, ${\displaystyle \frac{2\pi }{k}}$ is the period and $\varphi$ is the phase shift of the graph.
Here, in this case, $A=1$ , $k=1$ and $\varphi =0$ .
This means that amplitude of the function is $1$ , period is $2\pi$ and phase shift is zero.