Question

If $f\left(x\right)=\cos \left(\log x\right)$ , then show that $f\left(\frac{1}{x}\right).f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\frac{x}{y}+f\left(xy\right)\right]=0$

Answer 1

$f\left(x\right)\cos \left(\log x\right)$

$f\left(\frac{1}{x}\right)=\cos (\log \left(\frac{1}{x}\right)=\cos (\log 1-\log x)$

$=\cos (-\log x)=\cos \left(\log x\right)$

$f\left(\frac{1}{y}\right)=\cos \left(\log \left(\frac{1}{y}\right)\right)=\cos (\log 1-\log y)$

$=\cos (-\log y)=\cos \left(\log y\right)$

$f\left(\frac{x}{y}\right)=\cos \left(\log \left(\frac{x}{y}\right)\right)=\cos (\log x-\log y)$

$=\cos \left(logx\right)\cos \left(logy\right)+\sin \left(logx\right)\sin \left(logy\right)$

$f\left(xy\right)=\cos \left(\log \right(xy)=\cos (\log x+\log y)$

$=\cos \left(\log x\right)\cos \left(\log y\right)-\sin \left(\log x\right)\sin \left(\log y\right)$

Now,

$f\left(\frac{1}{x}\right).f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\right(\frac{x}{y})+f(xy\left)\right]$

$=\cos \left(\log x\right).\cos \left(\log y\right)-\frac{1}{2}\left[2\cos \right(\log x\left)\cos \right(\log y\left)\right]$

$=\cos \left(\log x\right)\cos \left(\log y\right)-\cos \left(\log x\right)\cos \left(\log y\right)$

$=0$