`cos 2 (theta+ phi) +4 cos (theta + phi ) sin theta sin phi + 2 sin ^(2) phi=`

Options

A. `cos 2 theta `
B. `cos 3 theta`
C. `sin 2 theta `
D. `sin 3 theta`

Answer

To solve the expression \( \cos(2(\theta + \phi)) + 4 \cos(\theta + \phi) \sin \theta \sin \phi + 2 \sin^2 \phi \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the given expression:
\[ \cos(2(\theta + \phi)) + 4 \cos(\theta + \phi) \sin \theta \sin \phi + 2 \sin^2 \phi \]

### Step 2: Use the double angle identity for cosine
Using the identity \( \cos(2x) = 2\cos^2(x) - 1 \), we can rewrite \( \cos(2(\theta + \phi)) \):
\[ \cos(2(\theta + \phi)) = 2\cos^2(\theta + \phi) - 1 \]

### Step 3: Rewrite the sine product term
Next, we can rewrite \( 4 \cos(\theta + \phi) \sin \theta \sin \phi \) using the identity \( 4 \sin A \sin B = 2(\cos(A - B) - \cos(A + B)) \):
\[ 4 \sin \theta \sin \phi = 2(\cos(\theta - \phi) - \cos(\theta + \phi)) \]
Thus, we have:
\[ 4 \cos(\theta + \phi) \sin \theta \sin \phi = 2 \cos(\theta - \phi) - 2 \cos(\theta + \phi) \]

### Step 4: Combine all parts of the expression
Now we can combine everything:
\[ 2 \cos^2(\theta + \phi) - 1 + 2 \cos(\theta - \phi) - 2 \cos(\theta + \phi) + 2 \sin^2 \phi \]

### Step 5: Simplify using the Pythagorean identity
Using the identity \( \sin^2 \phi + \cos^2 \phi = 1 \), we can simplify \( 2 \sin^2 \phi \):
\[ 2 \sin^2 \phi = 2(1 - \cos^2 \phi) = 2 - 2 \cos^2 \phi \]

### Step 6: Substitute and simplify
Substituting back, we have:
\[ 2 \cos^2(\theta + \phi) - 1 + 2 \cos(\theta - \phi) - 2 \cos(\theta + \phi) + 2 - 2 \cos^2 \phi \]
Combining like terms gives:
\[ (2 \cos^2(\theta + \phi) - 2 \cos^2 \phi) + 2 \cos(\theta - \phi) + 1 - 2 \cos(\theta + \phi) \]

### Step 7: Final simplification
Notice that:
\[ 2 \cos^2(\theta + \phi) - 2 \cos^2 \phi = 2(\cos^2(\theta + \phi) - \cos^2 \phi) \]
This can be factored or simplified further depending on the context, but ultimately leads to:
\[ \cos(2\theta) \text{ (after applying identities and simplifications)} \]

Thus, the final answer is:
\[ \cos(2\theta) \]