1. Tardigrade
  2. Question
  3. Mathematics
  4. If f(x) = cos (log x), then f ((1/x))f ((1/y))-(1/2) (f ((x/y))+f (xy)) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $f(x) = cos (log x)$, then $f \left(\frac{1}{x}\right)f \left(\frac{1}{y}\right)-\frac{1}{2} \left(f \left(\frac{x}{y}\right)+f \left(xy\right)\right)$ is equal to

Relations and Functions

Solution:

We have, $f \left(x\right) = cos\left(logx\right)$
$f \left(\frac{1}{x}\right)f \left(\frac{1}{y}\right)-\frac{1}{2}\left(f \left(\frac{x}{y}\right)+f \left(xy\right)\right)$
$=cos\left(log\frac{1}{x}\right)cos \left(log\frac{1}{y}\right)-\frac{1}{2}\left(cos\left(log\frac{x}{y}\right)+cos\left(log\, xy\right)\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)$
$-\frac{1}{2}\times2\, cos \left(\frac{log\frac{x}{y}+log\,xy}{2}\right)cos\left(\frac{log\frac{x}{y}-log\left(xy\right)}{2}\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)$
$-\left(cos\frac{\left(log\,x-log\,y+log\,x+log\,y\right)}{2}\right)\times$
$\left(cos\frac{log\,x-log\,y-log\,x-log\,y}{2}\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)-cos\left(log\,x\right)cos\left(log\,y\right)$
$=cos \left(log 1-log x\right)cos\left(log 1-log y\right)-cos\left(log\,x\right)cos \left(log\,y\right)$.
$=cos\left(log\,x\right)cos\left(log\,y\right)-cos\left(log\,x\right)cos\left(log\,y\right)$
$=0. \left(\because\, log\,1=0\right)$