Prove that
`sin (x + y ) sin (x -y) + sin ( y+ z) sin (y - z) + sin (z+x) sin (z-x) = 0.`

Answer

To prove that

\[ \sin(x+y) \sin(x-y) + \sin(y+z) \sin(y-z) + \sin(z+x) \sin(z-x) = 0, \]

we will use the product-to-sum identities for sine functions.

### Step 1: Apply the Product-to-Sum Identity
We know that:

\[ \sin(a+b) \sin(a-b) = \frac{1}{2} [\cos(2b) - \cos(2a)] \]

Using this identity, we can rewrite each term in the left-hand side (LHS):

1. For \(\sin(x+y) \sin(x-y)\):
\[ \sin(x+y) \sin(x-y) = \frac{1}{2} [\cos(2y) - \cos(2x)] \]

2. For \(\sin(y+z) \sin(y-z)\):
\[ \sin(y+z) \sin(y-z) = \frac{1}{2} [\cos(2z) - \cos(2y)] \]

3. For \(\sin(z+x) \sin(z-x)\):
\[ \sin(z+x) \sin(z-x) = \frac{1}{2} [\cos(2x) - \cos(2z)] \]

### Step 2: Combine the Terms
Now, substituting these results back into the LHS:

\[ \text{LHS} = \frac{1}{2} [\cos(2y) - \cos(2x)] + \frac{1}{2} [\cos(2z) - \cos(2y)] + \frac{1}{2} [\cos(2x) - \cos(2z)] \]

### Step 3: Simplify the Expression
Combining these terms gives:

\[ \text{LHS} = \frac{1}{2} [\cos(2y) - \cos(2x) + \cos(2z) - \cos(2y) + \cos(2x) - \cos(2z)] \]

Notice that \(\cos(2y)\) cancels out, \(\cos(2x)\) cancels out, and \(\cos(2z)\) cancels out:

\[ \text{LHS} = \frac{1}{2} [0] = 0 \]

### Conclusion
Thus, we have shown that:

\[ \sin(x+y) \sin(x-y) + \sin(y+z) \sin(y-z) + \sin(z+x) \sin(z-x) = 0 \]