If $f(x)=\cos(\log x)$, then $f(x)f(y)-\frac{1}{2}\left[f(\frac{x}{y})+f(xy)\right]$ has the value
Answer & explanation
Correct answer: option 4
$f(x)f(y)-\frac{1}{2}\left[f(\frac{x}{y})+f(xy)\right]=\cos(\log x)\cos(\log y)-\frac{1}{2}\left[\cos\left(\log(\frac{x}{y})\right)+\cos(\log(xy))\right]$
$=\cos(\log x)\cos(\log y)-\frac{1}{2}[\cos(\log x-\log y)+\cos(\log x+\log y)]$
$=\cos(\log x)\cos(\log y)-\frac{1}{2}[2\cos(\log x)\cos(\log y)]=0$