`sin (A+B) * sin (A-B)=`

Options

A. `sin^(2)A -cos^(2)B`
B. `cos^(2)A -sin^(2)B`
C. `sin^(2)A -sin^(2)B`
D. `cos^(2)A -cos^(2)B`

Answer

To solve the expression \( \sin(A+B) \cdot \sin(A-B) \), we can use the product-to-sum identities in trigonometry. Here’s a step-by-step solution:

### Step 1: Use the sine addition and subtraction formulas
The sine addition formula states:
\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]
The sine subtraction formula states:
\[ \sin(A-B) = \sin A \cos B - \cos A \sin B \]

### Step 2: Substitute the formulas into the expression
Now, substituting these formulas into the expression:
\[ \sin(A+B) \cdot \sin(A-B) = (\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B) \]

### Step 3: Recognize the difference of squares
This expression can be recognized as a difference of squares:
\[ (x+y)(x-y) = x^2 - y^2 \]
where \( x = \sin A \cos B \) and \( y = \cos A \sin B \).

### Step 4: Apply the difference of squares
Applying the difference of squares:
\[ \sin^2 A \cos^2 B - \cos^2 A \sin^2 B \]

### Step 5: Factor the expression
We can factor the expression further:
\[ = \sin^2 A \cos^2 B - \cos^2 A \sin^2 B = (\sin^2 A - \sin^2 B)(\cos^2 B + \cos^2 A) \]

### Final Result
Thus, the final expression simplifies to:
\[ \sin(A+B) \cdot \sin(A-B) = \frac{1}{2}(\cos(2B) - \cos(2A)) \]