If f(x)=cos(logeX), then find f (1/x)f(1/y) -1/2{f(xy)+f(x/y)} a) cos(...
Given ; f(x)= cos(loge x)= cos (log x) {since e is the base of log x}.
f(1/x)= cos(log 1/x)= cos(log x^-1) {since log x^m= m log x }.
= cos(-log x)=cos (log x) { since cos(-x)= cos(x) }.
Implies f(1/x)= cos(log x)= f(x).
similarly, f(1/y)= cos(log y) = f(y).
f(xy)= cos(log xy)= cos( log x +log y) {since log ab= log a+ log b }.
= cos(log x)*cos(log y) - sin(log x)*sin(log y) {since cos(A+B)= cos A* cos B- sin A* sin B }
f(x/y)= cos(log x/y)= cos(log x- log y) {since log a/b= log a- log b}.
= cos (log x)*cos(log y) + sin(log x)*sin(log y) {cos(A-B)= cos A*cos B + sin A*sin B }.
f(xy)+ f(x/y) = cos (log x)*cos(log y) - sin(log x)*sin(log y) + cos (log x)*cos(log y) + sin(log x)*sin(log y)
= 2*cos (log x)* cos (log y)
1/2{ f(xy)+ f(x/y) } = 1/2{ 2*cos (log x)* cos (log y) }
= cos (log x)* cos (log y).
f(1/x)f(1/y) - 1/2{ f(xy)+ f(x/y) } = cos (log x)* cos (log y) - cos (log x)* cos (log y).
= 0.
But, there is NO Answer in this question.