Question

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

Answer 1

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

$\to f\left(2/x\right)=\cos \left(\log c2/x\right)=\cos \left(\log x^2\right)=\cos (-\log x)$

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

$\to f\left(2/y\right)=\cos \left(\log x2/y\right)=\cos \left(\log y^{-1}\right)=\cos (-1\log y)$

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

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

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

$LHS$

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

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

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

$(\because \cos (\alpha -\beta )+\cos (\alpha +\beta )=2\cos \frac{\alpha +\beta }{2}.\cos \frac{\alpha -\beta }{2})$

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

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

$=0$

$=RHS$