What is the value of `cos [(180 - theta)//2] cos [(180 - 9theta)//2] + sin [(180 - 3theta)//2] sin [(180 - 13theta)//2]`?

Options

A. `sin 2theta sin 4theta`
B. `cos 2theta cos 6theta`
C. `sin 2theta sin 6theta`
D. `cos 2theta cos 4theta`

Answer

To solve the expression \( \cos \left( \frac{180 - \theta}{2} \right) \cos \left( \frac{180 - 9\theta}{2} \right) + \sin \left( \frac{180 - 3\theta}{2} \right) \sin \left( \frac{180 - 13\theta}{2} \right) \), we can use trigonometric identities to simplify it step by step.

### Step 1: Simplify using trigonometric identities
We start by using the identities:
- \( \cos(180^\circ - x) = -\cos(x) \)
- \( \sin(180^\circ - x) = \sin(x) \)

Thus, we can rewrite the expression:
\[ \cos \left( \frac{180 - \theta}{2} \right) = \cos \left( 90 - \frac{\theta}{2} \right) = \sin \left( \frac{\theta}{2} \right) \]
\[ \cos \left( \frac{180 - 9\theta}{2} \right) = \cos \left( 90 - \frac{9\theta}{2} \right) = \sin \left( \frac{9\theta}{2} \right) \]
\[ \sin \left( \frac{180 - 3\theta}{2} \right) = \sin \left( 90 - \frac{3\theta}{2} \right) = \cos \left( \frac{3\theta}{2} \right) \]
\[ \sin \left( \frac{180 - 13\theta}{2} \right) = \sin \left( 90 - \frac{13\theta}{2} \right) = \cos \left( \frac{13\theta}{2} \right) \]

Substituting these into the expression gives:
\[ \sin \left( \frac{\theta}{2} \right) \sin \left( \frac{9\theta}{2} \right) + \cos \left( \frac{3\theta}{2} \right) \cos \left( \frac{13\theta}{2} \right) \]

### Step 2: Use the product-to-sum identities
Next, we can apply the product-to-sum identities:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
\[ \cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \]

Applying these identities:
1. For \( \sin \left( \frac{\theta}{2} \right) \sin \left( \frac{9\theta}{2} \right) \):
\[ \sin \left( \frac{\theta}{2} \right) \sin \left( \frac{9\theta}{2} \right) = \frac{1}{2} \left[ \cos \left( \frac{\theta}{2} - \frac{9\theta}{2} \right) - \cos \left( \frac{\theta}{2} + \frac{9\theta}{2} \right) \right] = \frac{1}{2} \left[ \cos \left( -4\theta \right) - \cos \left( 5\theta \right) \right] = \frac{1}{2} \left[ \cos(4\theta) - \cos(5\theta) \right] \]

2. For \( \cos \left( \frac{3\theta}{2} \right) \cos \left( \frac{13\theta}{2} \right) \):
\[ \cos \left( \frac{3\theta}{2} \right) \cos \left( \frac{13\theta}{2} \right) = \frac{1}{2} \left[ \cos \left( \frac{3\theta}{2} - \frac{13\theta}{2} \right) + \cos \left( \frac{3\theta}{2} + \frac{13\theta}{2} \right) \right] = \frac{1}{2} \left[ \cos \left( -5\theta \right) + \cos \left( 8\theta \right) \right] = \frac{1}{2} \left[ \cos(5\theta) + \cos(8\theta) \right] \]

### Step 3: Combine the results
Now, we combine the two results:
\[ \frac{1}{2} \left[ \cos(4\theta) - \cos(5\theta) \right] + \frac{1}{2} \left[ \cos(5\theta) + \cos(8\theta) \right] \]

This simplifies to:
\[ \frac{1}{2} \left[ \cos(4\theta) + \cos(8\theta) \right] \]

### Conclusion
Thus, the final value of the original expression is:
\[ \frac{1}{2} \left[ \cos(4\theta) + \cos(8\theta) \right] \]