Find the value of `cos^(2)((pi)/(6) - (theta)/(2)) - sin^(2) ((pi)/(6)+(theta)/(2))`.

Answer

To find the value of \( \cos^2\left(\frac{\pi}{6} - \frac{\theta}{2}\right) - \sin^2\left(\frac{\pi}{6} + \frac{\theta}{2}\right) \), we can use the trigonometric identity:

\[ \cos^2 A - \sin^2 B = \cos(A + B) \cdot \cos(A - B) \]

### Step 1: Identify \( A \) and \( B \)

Let:
- \( A = \frac{\pi}{6} - \frac{\theta}{2} \)
- \( B = \frac{\pi}{6} + \frac{\theta}{2} \)

### Step 2: Calculate \( A + B \) and \( A - B \)

Now, we can calculate \( A + B \) and \( A - B \):

\[ A + B = \left(\frac{\pi}{6} - \frac{\theta}{2}\right) + \left(\frac{\pi}{6} + \frac{\theta}{2}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \]

\[ A - B = \left(\frac{\pi}{6} - \frac{\theta}{2}\right) - \left(\frac{\pi}{6} + \frac{\theta}{2}\right) = -\theta \]

### Step 3: Substitute into the identity

Now substitute \( A + B \) and \( A - B \) into the identity:

\[ \cos^2\left(\frac{\pi}{6} - \frac{\theta}{2}\right) - \sin^2\left(\frac{\pi}{6} + \frac{\theta}{2}\right) = \cos\left(\frac{\pi}{3}\right) \cdot \cos(-\theta) \]

### Step 4: Evaluate \( \cos\left(\frac{\pi}{3}\right) \) and \( \cos(-\theta) \)

We know that:

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

And since \( \cos(-\theta) = \cos(\theta) \):

\[ \cos(-\theta) = \cos(\theta) \]

### Step 5: Final expression

Putting it all together, we have:

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

### Final Answer

Thus, the final value is:

\[ \frac{1}{2} \cos(\theta) \]

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