To determine the phase shift of a trigonometric function, we begin by examining the argument inside the sine function. The phase shift of a function [tex]\(y = a \sin(bx - c)\)[/tex] is given by [tex]\(\frac{c}{b}\)[/tex].
Let's analyze each option to identify their phase shifts:
### Option A: [tex]\( y = 2 \sin\left(\frac{1}{2}x + \pi\right) \)[/tex]
For this function, [tex]\(a = 2\)[/tex], [tex]\(b = \frac{1}{2}\)[/tex], and [tex]\(c = -\pi\)[/tex]. The phase shift is:
[tex]\[
\text{Phase shift} = \frac{c}{b} = \frac{-\pi}{\frac{1}{2}} = -2\pi
\][/tex]
So, the phase shift is [tex]\(2\pi\)[/tex] to the left.
### Option B: [tex]\( y = 2 \sin(x - \pi) \)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = \pi\)[/tex]. The phase shift is:
[tex]\[
\text{Phase shift} = \frac{c}{b} = \frac{\pi}{1} = \pi
\][/tex]
So, the phase shift is [tex]\(\pi\)[/tex] to the right.
### Option C: [tex]\( y = 2 \sin(2x - \pi) \)[/tex]
In this case, [tex]\(a = 2\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = \pi\)[/tex]. The phase shift is:
[tex]\[
\text{Phase shift} = \frac{c}{b} = \frac{\pi}{2} = \frac{\pi}{2}
\][/tex]
So, the phase shift is [tex]\(\frac{\pi}{2}\)[/tex] to the right.
### Option D: [tex]\( y = 2 \sin\left(x + \frac{\pi}{2}\right) \)[/tex]
For this function, [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -\frac{\pi}{2}\)[/tex]. The phase shift is:
[tex]\[
\text{Phase shift} = \frac{c}{b} = \frac{-\frac{\pi}{2}}{1} = -\frac{\pi}{2}
\][/tex]
So, the phase shift is [tex]\(\frac{\pi}{2}\)[/tex] to the left.
### Conclusion
From the calculations, we see that the function [tex]\( y = 2 \sin(2x - \pi) \)[/tex] has a phase shift of [tex]\(\frac{\pi}{2}\)[/tex] to the right. Therefore, the correct answer is:
[tex]\[
\boxed{\text{C}}
\][/tex]
