Convert the products into sum or difference. If angles are given in degrees, evaluate from tables.
`cos (alpha + beta) cos (alpha - beta)`

Answer

To convert the product \( \cos(\alpha + \beta) \cos(\alpha - \beta) \) into a sum or difference, we can use a trigonometric identity. Here’s a step-by-step solution:

### Step 1: Write down the product
We start with the expression:
\[ \cos(\alpha + \beta) \cos(\alpha - \beta) \]

### Step 2: Use the product-to-sum identity
We can use the product-to-sum identity for cosine, which states:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
In our case, let \( A = \alpha + \beta \) and \( B = \alpha - \beta \).

### Step 3: Apply the identity
Using the identity, we rewrite our expression:
\[ \cos(\alpha + \beta) \cos(\alpha - \beta) = \frac{1}{2} [\cos((\alpha + \beta) + (\alpha - \beta)) + \cos((\alpha + \beta) - (\alpha - \beta))] \]

### Step 4: Simplify the angles
Now, we simplify the angles:
- For the first cosine term:
\[ (\alpha + \beta) + (\alpha - \beta) = 2\alpha \]
- For the second cosine term:
\[ (\alpha + \beta) - (\alpha - \beta) = 2\beta \]

### Step 5: Substitute back into the expression
Substituting back, we get:
\[ \cos(\alpha + \beta) \cos(\alpha - \beta) = \frac{1}{2} [\cos(2\alpha) + \cos(2\beta)] \]

### Final Result
Thus, the final result is:
\[ \cos(\alpha + \beta) \cos(\alpha - \beta) = \frac{1}{2} [\cos(2\alpha) + \cos(2\beta)] \]