Prove that
`sin ^(2) ((pi)/(8) + (theta)/(2)) - sin ^(2) ((pi)/(8) - (theta)/(2)) = sin ""(pi)/(4) sin theta.`

Answer

To prove that

\[ \sin^2\left(\frac{\pi}{8} + \frac{\theta}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{\theta}{2}\right) = \sin\left(\frac{\pi}{4}\right) \sin(\theta), \]

we will start by working on the left-hand side (LHS) of the equation.

### Step 1: Identify the LHS
Let
\[ LHS = \sin^2\left(\frac{\pi}{8} + \frac{\theta}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{\theta}{2}\right). \]

### Step 2: Use the identity for the difference of squares
We can use the identity \(a^2 - b^2 = (a+b)(a-b)\). Here, let \(A = \frac{\pi}{8} + \frac{\theta}{2}\) and \(B = \frac{\pi}{8} - \frac{\theta}{2}\). Thus, we have:

\[ LHS = \left(\sin\left(\frac{\pi}{8} + \frac{\theta}{2}\right) + \sin\left(\frac{\pi}{8} - \frac{\theta}{2}\right)\right) \left(\sin\left(\frac{\pi}{8} + \frac{\theta}{2}\right) - \sin\left(\frac{\pi}{8} - \frac{\theta}{2}\right)\right). \]

### Step 3: Simplify the first term
Using the sine addition formula, we have:

\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right). \]

Calculating \(A + B\) and \(A - B\):

\[ A + B = \left(\frac{\pi}{8} + \frac{\theta}{2}\right) + \left(\frac{\pi}{8} - \frac{\theta}{2}\right) = \frac{\pi}{4}, \]
\[ A - B = \left(\frac{\pi}{8} + \frac{\theta}{2}\right) - \left(\frac{\pi}{8} - \frac{\theta}{2}\right) = \theta. \]

Thus,

\[ \sin A + \sin B = 2 \sin\left(\frac{\pi}{4}\right) \cos\left(\frac{\theta}{2}\right). \]

### Step 4: Simplify the second term
Now for the second term, we have:

\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right). \]

Using the same calculations:

\[ \sin A - \sin B = 2 \cos\left(\frac{\pi}{4}\right) \sin\left(\frac{\theta}{2}\right). \]

### Step 5: Combine the results
Substituting these results back into the LHS:

\[ LHS = \left(2 \sin\left(\frac{\pi}{4}\right) \cos\left(\frac{\theta}{2}\right)\right) \left(2 \cos\left(\frac{\pi}{4}\right) \sin\left(\frac{\theta}{2}\right)\right). \]

### Step 6: Simplify further
Since \(\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\):

\[ LHS = 2 \cdot \frac{\sqrt{2}}{2} \cdot \cos\left(\frac{\theta}{2}\right) \cdot 2 \cdot \frac{\sqrt{2}}{2} \cdot \sin\left(\frac{\theta}{2}\right). \]

This simplifies to:

\[ LHS = 2 \cdot \frac{1}{2} \cdot \sin(\theta) = \sin(\theta). \]

### Step 7: Final result
Thus, we have:

\[ LHS = \sin\left(\frac{\pi}{4}\right) \sin(\theta). \]

Hence, we have proved that:

\[ \sin^2\left(\frac{\pi}{8} + \frac{\theta}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{\theta}{2}\right) = \sin\left(\frac{\pi}{4}\right) \sin(\theta). \]