Write the first four terms of the sequence whose nth term is given
`(-1)^(n-1) cos "" (npi)/(4)`

Answer

To find the first four terms of the sequence defined by the nth term \( t_n = (-1)^{n-1} \cos\left(\frac{n\pi}{4}\right) \), we will evaluate this expression for \( n = 1, 2, 3, \) and \( 4 \).

### Step 1: Calculate \( t_1 \)
For \( n = 1 \):
\[ t_1 = (-1)^{1-1} \cos\left(\frac{1\pi}{4}\right) = (-1)^0 \cos\left(\frac{\pi}{4}\right) \]
Since \( (-1)^0 = 1 \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ t_1 = 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 2: Calculate \( t_2 \)
For \( n = 2 \):
\[ t_2 = (-1)^{2-1} \cos\left(\frac{2\pi}{4}\right) = (-1)^1 \cos\left(\frac{\pi}{2}\right) \]
Since \( (-1)^1 = -1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ t_2 = -1 \cdot 0 = 0 \]

### Step 3: Calculate \( t_3 \)
For \( n = 3 \):
\[ t_3 = (-1)^{3-1} \cos\left(\frac{3\pi}{4}\right) = (-1)^2 \cos\left(\frac{3\pi}{4}\right) \]
Since \( (-1)^2 = 1 \) and \( \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \):
\[ t_3 = 1 \cdot \left(-\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]

### Step 4: Calculate \( t_4 \)
For \( n = 4 \):
\[ t_4 = (-1)^{4-1} \cos\left(\frac{4\pi}{4}\right) = (-1)^3 \cos(\pi) \]
Since \( (-1)^3 = -1 \) and \( \cos(\pi) = -1 \):
\[ t_4 = -1 \cdot (-1) = 1 \]

### Summary of the First Four Terms
The first four terms of the sequence are:
1. \( t_1 = \frac{\sqrt{2}}{2} \)
2. \( t_2 = 0 \)
3. \( t_3 = -\frac{\sqrt{2}}{2} \)
4. \( t_4 = 1 \)

Thus, the first four terms of the sequence are:
\[ \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, 1 \]