The range of `f(theta)=3cos^2theta-8sqrt(3) cos theta sin theta + 5 sin^2theta-7` is gven by

Options

A. `[-7,7]`
B. `[-10,4]`
C. `[-4,4]`
D. `[-10,7]`

Answer

To find the range of the function \( f(\theta) = 3\cos^2\theta - 8\sqrt{3} \cos\theta \sin\theta + 5\sin^2\theta - 7 \), we can follow these steps:

### Step 1: Rewrite the function using trigonometric identities
We know that:
\[ \cos^2\theta = \frac{1 + \cos 2\theta}{2} \]
\[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \]
\[ \sin 2\theta = 2\sin\theta\cos\theta \]
Using these identities, we can rewrite \( f(\theta) \).

### Step 2: Substitute the identities into the function
Substituting the identities into \( f(\theta) \):
\[ f(\theta) = 3\left(\frac{1 + \cos 2\theta}{2}\right) - 8\sqrt{3}\left(\frac{\sin 2\theta}{2}\right) + 5\left(\frac{1 - \cos 2\theta}{2}\right) - 7 \]
This simplifies to:
\[ f(\theta) = \frac{3 + 3\cos 2\theta - 4\sqrt{3}\sin 2\theta + 5 - 5\cos 2\theta - 14}{2} \]

### Step 3: Combine like terms
Combining the terms gives:
\[ f(\theta) = \frac{-2\cos 2\theta - 4\sqrt{3}\sin 2\theta - 6}{2} \]
This simplifies to:
\[ f(\theta) = -\cos 2\theta - 2\sqrt{3}\sin 2\theta - 3 \]

### Step 4: Express in the form \( a\cos x + b\sin x \)
We rewrite \( f(\theta) \) in the form \( A\cos(2\theta + \phi) + C \):
\[ f(\theta) = -2\sqrt{3}\sin 2\theta - 2\cos 2\theta - 3 \]
Let \( A = -2 \) and \( B = -2\sqrt{3} \).

### Step 5: Find the maximum and minimum values
The maximum and minimum values of \( A\cos x + B\sin x \) can be found using:
\[ R = \sqrt{A^2 + B^2} \]
Calculating \( R \):
\[ R = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

### Step 6: Determine the range of \( f(\theta) \)
The maximum value of \( -2\cos 2\theta - 4\sqrt{3}\sin 2\theta \) is \( 4 \) and the minimum value is \( -4 \). Therefore:
- Maximum value of \( f(\theta) = -3 + 4 = 1 \)
- Minimum value of \( f(\theta) = -3 - 4 = -7 \)

### Final Range
Thus, the range of \( f(\theta) \) is:
\[ [-7, 1] \]