Practicing Success
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 |
1 $\frac{1}{2}$ -2 0 |
0 |
$f(x)f(y)-\frac{1}{2}\left[f(\frac{x}{y})+f(xy)\right]=\cos(\log x)\cos(\log y)-\frac{1}{2}\left[\cos\left(\log(\frac{x}{y})\right)+\cos(\log(xy))\right]$ $=\cos(\log x)\cos(\log y)-\frac{1}{2}[\cos(\log x-\log y)+\cos(\log x+\log y)]$ $=\cos(\log x)\cos(\log y)-\frac{1}{2}[2\cos(\log x)\cos(\log y)]=0$ |