What is the range of f(x) = cos2x-sin2x ?

Options

A. [2, 4]
B. [-1,1]
C. `[-sqrt(2).sqrt(2)]`
D. `(-sqrt(2)sqrt(2))`

Answer

To find the range of the function \( f(x) = \cos(2x) - \sin(2x) \), we can follow these steps:

### Step 1: Rewrite the function
We can express the function in a different form to make it easier to analyze. We can use the identity for cosine of a sum:
\[ f(x) = \cos(2x) - \sin(2x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos(2x) - \frac{1}{\sqrt{2}} \sin(2x) \right) \]
This can be rewritten as:
\[ f(x) = \sqrt{2} \left( \cos(2x + \frac{\pi}{4}) \right) \]
This is because \( \frac{1}{\sqrt{2}} \) corresponds to the cosine and sine of \( 45^\circ \) or \( \frac{\pi}{4} \).

### Step 2: Determine the maximum and minimum values
The cosine function oscillates between -1 and 1. Therefore, the maximum and minimum values of \( f(x) \) can be determined as follows:
- Maximum value of \( f(x) \):
\[ \text{Max} = \sqrt{2} \cdot 1 = \sqrt{2} \]
- Minimum value of \( f(x) \):
\[ \text{Min} = \sqrt{2} \cdot (-1) = -\sqrt{2} \]

### Step 3: Define the range
From the maximum and minimum values calculated, we can define the range of the function:
\[ \text{Range of } f(x) = [-\sqrt{2}, \sqrt{2}] \]

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

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