`cos^2 (pi/6 + theta) - sin^2 (pi/6 - theta) =`

Options

A. `1/2 cos 2 theta `
B. 0
C. `-1/2 cos 2 theta `
D. `1/2`

Answer

To solve the expression \( \cos^2\left(\frac{\pi}{6} + \theta\right) - \sin^2\left(\frac{\pi}{6} - \theta\right) \), we can use the trigonometric identity for the difference of squares.

### Step-by-step Solution:

1. **Identify the Expression**:
\[ \cos^2\left(\frac{\pi}{6} + \theta\right) - \sin^2\left(\frac{\pi}{6} - \theta\right) \]

2. **Use the Identity**:
Recall the identity:
\[ \cos^2 A - \sin^2 B = \cos A \cos B + \sin A \sin B \]
Here, we can rewrite the expression using the identity:
\[ \cos^2 A - \sin^2 B = \cos\left(\frac{\pi}{6} + \theta\right)\cos\left(\frac{\pi}{6} - \theta\right) + \sin\left(\frac{\pi}{6} + \theta\right)\sin\left(\frac{\pi}{6} - \theta\right) \]

3. **Set A and B**:
Let \( A = \frac{\pi}{6} + \theta \) and \( B = \frac{\pi}{6} - \theta \).

4. **Apply the Cosine and Sine Addition Formulas**:
Using the cosine and sine addition formulas:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

5. **Calculate \( \cos A \) and \( \sin A \)**:
\[ \cos A = \cos\left(\frac{\pi}{6} + \theta\right) \]
\[ \sin A = \sin\left(\frac{\pi}{6} + \theta\right) \]
\[ \cos B = \cos\left(\frac{\pi}{6} - \theta\right) \]
\[ \sin B = \sin\left(\frac{\pi}{6} - \theta\right) \]

6. **Combine the Results**:
Substitute back into the expression:
\[ \cos^2\left(\frac{\pi}{6} + \theta\right) - \sin^2\left(\frac{\pi}{6} - \theta\right) = \cos\left(\frac{\pi}{3}\right) \cos(2\theta) \]

7. **Evaluate \( \cos\left(\frac{\pi}{3}\right) \)**:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

8. **Final Expression**:
Therefore, the final expression simplifies to:
\[ \frac{1}{2} \cos(2\theta) \]

### Final Answer:
\[ \cos^2\left(\frac{\pi}{6} + \theta\right) - \sin^2\left(\frac{\pi}{6} - \theta\right) = \frac{1}{2} \cos(2\theta) \]