Simplify `sin(A+45^(@))sin(A-45^(@))`.

Answer

To simplify the expression \( \sin(A + 45^\circ) \sin(A - 45^\circ) \), we can use the product-to-sum identities in trigonometry. Here’s a step-by-step solution:

### Step 1: Use the Product-to-Sum Identity
We can use the product-to-sum identity for sine functions:
\[ \sin x \sin y = \frac{1}{2} [\cos(x - y) - \cos(x + y)] \]
In our case, let \( x = A + 45^\circ \) and \( y = A - 45^\circ \).

### Step 2: Calculate \( x - y \) and \( x + y \)
Now, we calculate:
- \( x - y = (A + 45^\circ) - (A - 45^\circ) = 90^\circ \)
- \( x + y = (A + 45^\circ) + (A - 45^\circ) = 2A \)

### Step 3: Substitute into the Identity
Substituting these values into the product-to-sum formula gives:
\[ \sin(A + 45^\circ) \sin(A - 45^\circ) = \frac{1}{2} [\cos(90^\circ) - \cos(2A)] \]

### Step 4: Simplify the Expression
We know that \( \cos(90^\circ) = 0 \), so:
\[ \sin(A + 45^\circ) \sin(A - 45^\circ) = \frac{1}{2} [0 - \cos(2A)] = -\frac{1}{2} \cos(2A) \]

### Final Result
Thus, the simplified form of \( \sin(A + 45^\circ) \sin(A - 45^\circ) \) is:
\[ -\frac{1}{2} \cos(2A) \]

---