If `f(x)=sin(logx)` then `f(xy)+f(x/y)-2f(x)cos(logy)=` (A) `cos(logx)` (B) `sin(logy)` (C) `cos(log(xy))` (D) 0

Answer

To solve the problem, we start with the function defined as \( f(x) = \sin(\log x) \). We need to evaluate the expression \( f(xy) + f\left(\frac{x}{y}\right) - 2f(x)\cos(\log y) \).

### Step 1: Calculate \( f(xy) \)

Using the definition of the function:
\[ f(xy) = \sin(\log(xy)) = \sin(\log x + \log y) \]
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
we can rewrite this as:
\[ f(xy) = \sin(\log x) \cos(\log y) + \cos(\log x) \sin(\log y) \]

### Step 2: Calculate \( f\left(\frac{x}{y}\right) \)

Now we calculate \( f\left(\frac{x}{y}\right) \):
\[ f\left(\frac{x}{y}\right) = \sin\left(\log\left(\frac{x}{y}\right)\right) = \sin(\log x - \log y) \]
Using the sine subtraction formula:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
we can rewrite this as:
\[ f\left(\frac{x}{y}\right) = \sin(\log x) \cos(\log y) - \cos(\log x) \sin(\log y) \]

### Step 3: Substitute into the expression

Now, we substitute \( f(xy) \) and \( f\left(\frac{x}{y}\right) \) into the original expression:
\[ f(xy) + f\left(\frac{x}{y}\right) - 2f(x)\cos(\log y) \]
Substituting the values we found:
\[ = \left(\sin(\log x) \cos(\log y) + \cos(\log x) \sin(\log y)\right) + \left(\sin(\log x) \cos(\log y) - \cos(\log x) \sin(\log y)\right) - 2\sin(\log x)\cos(\log y) \]

### Step 4: Simplify the expression

Combining the terms:
\[ = \sin(\log x) \cos(\log y) + \sin(\log x) \cos(\log y) + \cos(\log x) \sin(\log y) - \cos(\log x) \sin(\log y) - 2\sin(\log x)\cos(\log y) \]
This simplifies to:
\[ = 2\sin(\log x)\cos(\log y) + 0 - 2\sin(\log x)\cos(\log y) = 0 \]

### Conclusion

Thus, the final result is:
\[ f(xy) + f\left(\frac{x}{y}\right) - 2f(x)\cos(\log y) = 0 \]

### Final Answer

The answer is \( \boxed{0} \).