Evaluate `sin{ n pi+(-1)^(n) (pi)/(4)`, where n is an integer.

Answer

To evaluate the expression \( \sin\left(n \pi + (-1)^{n} \frac{\pi}{4}\right) \), we can break it down step by step.

### Step 1: Understand the Expression
The expression consists of two parts: \( n \pi \) and \( (-1)^{n} \frac{\pi}{4} \). The term \( (-1)^{n} \) will determine whether we are adding or subtracting \( \frac{\pi}{4} \) based on whether \( n \) is even or odd.

### Step 2: Evaluate for Even and Odd \( n \)
1. **If \( n \) is even**:
- \( (-1)^{n} = 1 \)
- Thus, the expression becomes:
\[ \sin\left(n \pi + \frac{\pi}{4}\right) \]
- Using the sine addition formula, we can rewrite this as:
\[ \sin\left(n \pi + \frac{\pi}{4}\right) = \sin(n \pi)\cos\left(\frac{\pi}{4}\right) + \cos(n \pi)\sin\left(\frac{\pi}{4}\right) \]
- Since \( \sin(n \pi) = 0 \) and \( \cos(n \pi) = (-1)^{n} = 1 \) (for even \( n \)):
\[ = 0 \cdot \cos\left(\frac{\pi}{4}\right) + 1 \cdot \sin\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) \]
- Therefore, we have:
\[ \sin\left(n \pi + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

2. **If \( n \) is odd**:
- \( (-1)^{n} = -1 \)
- Thus, the expression becomes:
\[ \sin\left(n \pi - \frac{\pi}{4}\right) \]
- Again using the sine subtraction formula:
\[ \sin\left(n \pi - \frac{\pi}{4}\right) = \sin(n \pi)\cos\left(-\frac{\pi}{4}\right) - \cos(n \pi)\sin\left(-\frac{\pi}{4}\right) \]
- Here, \( \sin(n \pi) = 0 \) and \( \cos(n \pi) = (-1)^{n} = -1 \) (for odd \( n \)):
\[ = 0 \cdot \cos\left(-\frac{\pi}{4}\right) - (-1) \cdot \sin\left(-\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) \]
- Therefore, we have:
\[ \sin\left(n \pi - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Conclusion
In both cases (whether \( n \) is even or odd), we find that:
\[ \sin\left(n \pi + (-1)^{n} \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the value of \( \sin\left(n \pi + (-1)^{n} \frac{\pi}{4}\right) \) is \( \frac{1}{\sqrt{2}} \).

---