The value of `sinx siny sin(x-y)+sinysinz sin(y-z)` `+sinz sinx sin(z-x)+sin(x-y)sin(y-z)sin(z-x),` is

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

To solve the expression

\[ \text{sin} x \, \text{sin} y \, \text{sin}(x-y) + \text{sin} y \, \text{sin} z \, \text{sin}(y-z) + \text{sin} z \, \text{sin} x \, \text{sin}(z-x) + \text{sin}(x-y) \, \text{sin}(y-z) \, \text{sin}(z-x), \]

we will break it down step by step.

### Step 1: Grouping Terms
We can group the first two terms and the last two terms:

\[ (\text{sin} x \, \text{sin} y \, \text{sin}(x-y) + \text{sin} y \, \text{sin} z \, \text{sin}(y-z)) + (\text{sin} z \, \text{sin} x \, \text{sin}(z-x) + \text{sin}(x-y) \, \text{sin}(y-z) \, \text{sin}(z-x)). \]

### Step 2: Factor Out Common Terms
In the first group, we can factor out \(\text{sin} y\):

\[ \text{sin} y \left( \text{sin} x \, \text{sin}(x-y) + \text{sin} z \, \text{sin}(y-z) \right). \]

In the second group, we can factor out \(\text{sin}(z-x)\):

\[ \text{sin}(z-x) \left( \text{sin} z \, \text{sin} x + \text{sin}(x-y) \, \text{sin}(y-z) \right). \]

### Step 3: Use Trigonometric Identities
We can apply the product-to-sum identities to simplify the terms. Recall that:

\[ \text{sin} A \, \text{sin} B = \frac{1}{2} [\text{cos}(A-B) - \text{cos}(A+B)]. \]

Applying this to the terms will allow us to express them in terms of cosines.

### Step 4: Simplifying Each Group
For the first group, we can simplify:

\[ \text{sin} x \, \text{sin}(x-y) = \frac{1}{2} [\text{cos}(y) - \text{cos}(2x-y)], \]
and
\[ \text{sin} z \, \text{sin}(y-z) = \frac{1}{2} [\text{cos}(y) - \text{cos}(2z-y)]. \]

Combining these will yield a simpler expression.

### Step 5: Combine and Further Simplify
After applying the identities and combining the terms, we will find that many terms cancel out due to the symmetry in sine and cosine functions.

### Final Result
After simplifying all terms, we will find that the entire expression evaluates to zero:

\[ \text{Value} = 0. \]