Convert the products into sum or difference. If angles are given in degrees, evaluate from tables.
`sin ""(A +B)/(2) cos ""(A -B)/(2)`

Answer

To solve the problem of converting the product \( \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \) into a sum or difference, we can use the product-to-sum identities. Here’s the step-by-step solution:

### Step 1: Identify the Product-to-Sum Identity
We know from trigonometric identities that:
\[ \sin C \cos D = \frac{1}{2} \left( \sin(C + D) + \sin(C - D) \right) \]
In our case, let \( C = \frac{A + B}{2} \) and \( D = \frac{A - B}{2} \).

### Step 2: Apply the Identity
Using the identity, we can write:
\[ \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = \frac{1}{2} \left( \sin\left(\frac{A + B}{2} + \frac{A - B}{2}\right) + \sin\left(\frac{A + B}{2} - \frac{A - B}{2}\right) \right) \]

### Step 3: Simplify the Angles
Now, we simplify the angles inside the sine functions:
1. For \( \frac{A + B}{2} + \frac{A - B}{2} \):
\[ = \frac{(A + B) + (A - B)}{2} = \frac{2A}{2} = A \]

2. For \( \frac{A + B}{2} - \frac{A - B}{2} \):
\[ = \frac{(A + B) - (A - B)}{2} = \frac{2B}{2} = B \]

### Step 4: Substitute Back
Substituting these results back into our equation gives:
\[ \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = \frac{1}{2} \left( \sin A + \sin B \right) \]

### Final Result
Thus, we conclude that:
\[ \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = \frac{1}{2} \left( \sin A + \sin B \right) \]