`cos 2 theta cos 2 phi + sin^(2) (theta - phi) - sin^(2) (theta + phi)` is equal to

Options

A. `sin 2 (theta + phi)`
B. `cos 2 (theta + phi)`
C. `sin 2(thea -phi)`
D. `cos 2 (theta -phi)`

Answer

To solve the expression \( \cos 2\theta \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \), we will simplify it step by step.

### Step 1: Recognize the identity for the difference of squares
We can rewrite \( \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \) using the identity for the difference of squares:
\[ a^2 - b^2 = (a + b)(a - b) \]
Here, let \( a = \sin(\theta - \phi) \) and \( b = \sin(\theta + \phi) \).

### Step 2: Apply the identity
Thus, we have:
\[ \sin^2(\theta - \phi) - \sin^2(\theta + \phi) = (\sin(\theta - \phi) + \sin(\theta + \phi))(\sin(\theta - \phi) - \sin(\theta + \phi)) \]

### Step 3: Use the sine addition and subtraction formulas
Using the sine addition and subtraction formulas:
\[ \sin(\theta + \phi) = \sin\theta \cos\phi + \cos\theta \sin\phi \]
\[ \sin(\theta - \phi) = \sin\theta \cos\phi - \cos\theta \sin\phi \]
Now, substituting these into our expression:
\[ \sin(\theta - \phi) + \sin(\theta + \phi) = 2\sin\theta \cos\phi \]
\[ \sin(\theta - \phi) - \sin(\theta + \phi) = -2\cos\theta \sin\phi \]

### Step 4: Substitute back into the expression
Now substituting these results back into our expression:
\[ \sin^2(\theta - \phi) - \sin^2(\theta + \phi) = (2\sin\theta \cos\phi)(-2\cos\theta \sin\phi) = -4\sin\theta \cos\theta \sin\phi \cos\phi \]

### Step 5: Combine with the cosine product
Now, we combine this with the \( \cos 2\theta \cos 2\phi \):
\[ \cos 2\theta \cos 2\phi - 4\sin\theta \cos\theta \sin\phi \cos\phi \]

### Step 6: Use product-to-sum identities
Recall the product-to-sum identities:
\[ \cos A \cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B)) \]
Thus, we can express \( \cos 2\theta \cos 2\phi \):
\[ \cos 2\theta \cos 2\phi = \frac{1}{2}(\cos(2\theta + 2\phi) + \cos(2\theta - 2\phi)) \]

### Step 7: Final expression
Putting everything together, we get:
\[ \frac{1}{2}(\cos(2\theta + 2\phi) + \cos(2\theta - 2\phi)) - 4\sin\theta \cos\theta \sin\phi \cos\phi \]

### Conclusion
Thus, the final simplified form of the expression \( \cos 2\theta \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \) is:
\[ \cos(2\theta + 2\phi) + \cos(2\theta - 2\phi) - 4\sin\theta \cos\theta \sin\phi \cos\phi \]