Question

If f(x)=cos(log x), then prove that

f(1/x).f(1/y)-1/2(f(x/y)+f(xy))=0

Answer 1

y = f(x) = cos(log x)
f(1/x) = cos(log 1/x) = cos(- log x) = cos(log x)
f(1/y) = cos(log 1/y) = cos(- log y) = cos(log y)
f(x/y) = cos(log x/y) = cos(log x - log y) = cos(logx)cos(logy) + sin(logx)sin(logy)
f(xy) = cos(log xy) = cos(log x + log y) = cos(logx)cos(logy) - sin(logx)sin(logy)
Now,
f(1/x)f(1/y) - 1/2[f(x/y) + f(xy)] =
cos(log x) cos(log y) - 1/2[cos(logx)cos(logy) + sin(logx)sin(logy) + cos(logx)cos(logy) - sin(logx)sin(logy)] =
cos(log x) cos(log y) - 1/2[cos(logx)cos(logy) + cos(logx)cos(logy)] =
cos(log x) cos(log y) - cos(logx)cos(logy) =
cos(log x) cos(log y) - cos(logx)cos(logy) =
0