What is the amplitude, period, phase shift, and vertical displacement of $y = \sin x - 1$ ?

Hint: First, the general equation of wave has to be known. Since the sinusoidal equation is given, the general equation should also be sinusoidal. The amplitude is the highest displacement of the wave that we can get from the general equation. Also, the period, phase shift, and vertical displacement are noted in that equation. The given equation has to be equated with the general one to find the values of required amplitude, period, phase, and vertical displacement.

Formula used:
The general equation: $y = a\sin (px - q) + r$
Where, $a = $ amplitude
$r = $ the vertical shift.
$\dfrac{{2\pi }}{p} = $ period.
$q = $ the phase shift.

Complete answer:
The given equation is, $y = \sin x - 1$
Clearly, it is a sinusoidal equation.
The general equation of wave is to take here, $y = a\sin (px - q) + r$
Where, $a = $ amplitude
$r = $ the vertical shift.
$\dfrac{{2\pi }}{p} = $ period.
$q = $ the phase shift.
If we write $y = \sin x - 1$ as $y = a\sin (px - q) + r$, it will be like:
$y = 1.\sin (1.x - 0) + ( - 1)$
By equating with the general equation we get from $y = 1.\sin (1.x - 0) + ( - 1)$,
$a = 1$, $p = 1$, $q = 0$ , $r = - 1$
Therefore, the amplitude $a = 1$
The period,$\dfrac{{2\pi }}{p} = \dfrac{{2\pi }}{1} = 2\pi $
The phase shift, $q = 0$
The vertical displacement, $r = - 1$

Note:
$y = \sin x - 1$ - This function is the same as $y = \sin x$ except it starts from $( - 1)$ rather than zero.
Also, the amplitude$\left( {a = 1} \right)$ is the same as that of the function $y = \sin x$ and the period$(2\pi )$ is similar to that of the function $y = \sin x$.
For the equation when there is $\sin x,\cos x,\sec x$ we take the period $2\pi $ and for the equation when there is $\tan x,\cot x$ we take the period $\pi $.
The phase shift we get here is, $q = 0$which means there is no shift in left or right.
The vertical displacement, $r = - 1$, which means the value of the function at $x = 0$.