The range of `f(x)=cosx-sinx` is

Options

A. `[-1,1]`
B. `(-1,2)`
C. `[-(pi)/(2),(pi)/(2)]`
D. `[-sqrt(2),sqrt(2)]`

Explanation

To find the range of the function \( f(x) = \cos x - \sin x \), we can use the properties of trigonometric functions and the concept of combining them in a specific way.

### Step-by-step Solution:

1. **Identify the Function**:
We have the function \( f(x) = \cos x - \sin x \).

2. **Rewrite the Function**:
We can express \( f(x) \) in the form of \( a \cos x + b \sin x \). Here, \( a = 1 \) and \( b = -1 \).

3. **Use the Range Formula**:
The range of a function of the form \( a \cos x + b \sin x \) can be determined using the formula:
\[ R = \sqrt{a^2 + b^2} \]
Therefore, we calculate:
\[ R = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

4. **Determine the Maximum and Minimum Values**:
The maximum value of \( f(x) \) is \( \sqrt{2} \) and the minimum value is \( -\sqrt{2} \). Thus, we can express the range of \( f(x) \) as:
\[ -\sqrt{2} \leq f(x) \leq \sqrt{2} \]

5. **Conclusion**:
Therefore, the range of the function \( f(x) = \cos x - \sin x \) is:
\[ \text{Range} = [-\sqrt{2}, \sqrt{2}] \]

### Final Answer:
The range of \( f(x) = \cos x - \sin x \) is \( [-\sqrt{2}, \sqrt{2}] \).

---