Classify the following functions for being even or odd:
` sin x +cos x`

Answer

To classify the function \( f(x) = \sin x + \cos x \) as even, odd, or neither, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \sin x + \cos x \).

### Step 2: Find \( f(-x) \)
We need to compute \( f(-x) \):
\[ f(-x) = \sin(-x) + \cos(-x) \]

### Step 3: Use properties of sine and cosine
Recall the properties of sine and cosine:
- \( \sin(-x) = -\sin(x) \)
- \( \cos(-x) = \cos(x) \)

Using these properties, we can rewrite \( f(-x) \):
\[ f(-x) = -\sin(x) + \cos(x) \]

### Step 4: Compare \( f(-x) \) with \( f(x) \)
Now we compare \( f(-x) \) with \( f(x) \):
- We have \( f(x) = \sin x + \cos x \)
- We found \( f(-x) = -\sin x + \cos x \)

### Step 5: Check for even function
A function is even if \( f(-x) = f(x) \). Here:
\[ f(-x) \neq f(x) \]
Thus, \( f(x) \) is not an even function.

### Step 6: Check for odd function
A function is odd if \( f(-x) = -f(x) \). We need to compute \( -f(x) \):
\[ -f(x) = -(\sin x + \cos x) = -\sin x - \cos x \]

Now, we compare \( f(-x) \) with \( -f(x) \):
\[ f(-x) = -\sin x + \cos x \quad \text{and} \quad -f(x) = -\sin x - \cos x \]
Since \( f(-x) \neq -f(x) \), the function is not odd.

### Conclusion
Since \( f(x) \) is neither even nor odd, we conclude that:
\[ f(x) = \sin x + \cos x \text{ is neither even nor odd.} \]