`cos2thetacos2phi+sin^2(theta-phi)-sin^2(theta+phi)=`

Options

A. `sin2(theta+phi)`
B. `cos2(theta+phi)`
C. `sin2(theta-phi)`
D. `cos2(theta-phi)`

Explanation

To solve the expression \( \cos 2\theta \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the given expression:
\[ \cos 2\theta \cos 2\phi + \sin^2(\theta - \phi) - \sin^2(\theta + \phi) \]

### Step 2: Use the identity for sine squares
Recall the identity for the difference of squares:
\[ \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) \]
Here, let \( A = \theta + \phi \) and \( B = \theta - \phi \). Thus, we can rewrite:
\[ \sin^2(\theta - \phi) - \sin^2(\theta + \phi) = -(\sin^2(\theta + \phi) - \sin^2(\theta - \phi)) \]

### Step 3: Apply the sine addition and subtraction formulas
Using the sine addition and subtraction formulas:
\[ \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi \]
\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi \]
We can express:
\[ \sin^2(\theta + \phi) = (\sin \theta \cos \phi + \cos \theta \sin \phi)^2 \]
\[ \sin^2(\theta - \phi) = (\sin \theta \cos \phi - \cos \theta \sin \phi)^2 \]

### Step 4: Expand the squares
Expanding both squares:
\[ \sin^2(\theta + \phi) = \sin^2 \theta \cos^2 \phi + 2 \sin \theta \cos \theta \sin \phi \cos \phi + \cos^2 \theta \sin^2 \phi \]
\[ \sin^2(\theta - \phi) = \sin^2 \theta \cos^2 \phi - 2 \sin \theta \cos \theta \sin \phi \cos \phi + \cos^2 \theta \sin^2 \phi \]

### Step 5: Substitute back into the expression
Now substituting back:
\[ \cos 2\theta \cos 2\phi + \left(\sin^2 \theta \cos^2 \phi - 2 \sin \theta \cos \theta \sin \phi \cos \phi + \cos^2 \theta \sin^2 \phi\right) - \left(\sin^2 \theta \cos^2 \phi + 2 \sin \theta \cos \theta \sin \phi \cos \phi + \cos^2 \theta \sin^2 \phi\right) \]

### Step 6: Simplify the expression
Notice that the terms involving \( \sin^2 \theta \cos^2 \phi \) and \( \cos^2 \theta \sin^2 \phi \) will cancel out:
\[ \cos 2\theta \cos 2\phi - 4 \sin \theta \cos \theta \sin \phi \cos \phi \]

### Step 7: Use double angle identities
Using the double angle identity:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \quad \text{and} \quad \sin 2\phi = 2 \sin \phi \cos \phi \]
We can rewrite:
\[ -4 \sin \theta \cos \theta \sin \phi \cos \phi = -2 \sin 2\theta \sin 2\phi \]

### Step 8: Final expression
Thus, we can express the entire expression as:
\[ \cos 2\theta \cos 2\phi - 2 \sin 2\theta \sin 2\phi \]

### Step 9: Use the cosine of sum identity
Using the identity:
\[ \cos A \cos B - \sin A \sin B = \cos(A + B) \]
We have:
\[ \cos(2\theta + 2\phi) \]

### Final Answer
The final result is:
\[ \cos(2\theta + 2\phi) \]

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