Find the value of ` cos {npi +(-1) ^(n) (pi)/(3)} `

Answer

To solve the problem \( \cos(n\pi + (-1)^n \frac{\pi}{3}) \), we can break it down into two cases based on whether \( n \) is odd or even.

### Step 1: Identify the cases for \( n \)
- **Case 1:** \( n \) is odd (e.g., \( n = 1, 3, 5, \ldots \))
- **Case 2:** \( n \) is even (e.g., \( n = 0, 2, 4, \ldots \))

### Step 2: Evaluate for odd \( n \)
When \( n \) is odd:
- \( (-1)^n = -1 \)
- Therefore, the expression becomes:
\[ \cos(n\pi - \frac{\pi}{3}) \]
- Using the cosine subtraction formula:
\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]
- Here, \( a = n\pi \) and \( b = \frac{\pi}{3} \):
\[ \cos(n\pi - \frac{\pi}{3}) = \cos(n\pi)\cos(\frac{\pi}{3}) + \sin(n\pi)\sin(\frac{\pi}{3}) \]
- Since \( \cos(n\pi) = -1 \) (for odd \( n \)) and \( \sin(n\pi) = 0 \):
\[ \cos(n\pi - \frac{\pi}{3}) = -1 \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{3}}{2} = -\frac{1}{2} \]

### Step 3: Evaluate for even \( n \)
When \( n \) is even:
- \( (-1)^n = 1 \)
- Therefore, the expression becomes:
\[ \cos(n\pi + \frac{\pi}{3}) \]
- Using the cosine addition formula:
\[ \cos(a + b) = \cos a \cos b - \sin a \sin b \]
- Here, \( a = n\pi \) and \( b = \frac{\pi}{3} \):
\[ \cos(n\pi + \frac{\pi}{3}) = \cos(n\pi)\cos(\frac{\pi}{3}) - \sin(n\pi)\sin(\frac{\pi}{3}) \]
- Since \( \cos(n\pi) = 1 \) (for even \( n \)) and \( \sin(n\pi) = 0 \):
\[ \cos(n\pi + \frac{\pi}{3}) = 1 \cdot \frac{1}{2} - 0 \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \]

### Conclusion
- If \( n \) is odd, \( \cos(n\pi + (-1)^n \frac{\pi}{3}) = -\frac{1}{2} \)
- If \( n \) is even, \( \cos(n\pi + (-1)^n \frac{\pi}{3}) = \frac{1}{2} \)

### Final Answer
\[ \cos(n\pi + (-1)^n \frac{\pi}{3}) = \begin{cases} -\frac{1}{2} & \text{if } n \text{ is odd} \\ \frac{1}{2} & \text{if } n \text{ is even} \end{cases} \]