Question

If f(x)=cos (log x ) then f(x) f (y)-1/2 {f(x/y)+f(xy)} then what is the value of this

Answer 1

F(x)=cos(logx)
∴, f(y)=cos(logy), f(x/y)=coslog(x/y), f(xy)=coslog(xy)
∴, f(x)f(y)-1/2{f(x/y)+f(xy)}
=cos(logx)cos(logy)-1/2{coslog(x/y)+coslog(xy)}
=cos(logx)cos(logy)-1/2[2cos{log(x/y)+log(xy)}/2cos{log(x/y)-log(xy)}/2]
=cos(logx)cos(logy)-cos{(logx-logy+logx+logy)/2}cos{(logx-logy-logx-logy)/2}
=cos(logx)cos(logy)-cos{(2logx)/2}cos{(-2logy)/2}
=cos(logx)cos(logy)-cos(logx)cos(logy) [∵, cos(-logy)=cos(logy)]
=0