Question

If f(x) = cos (log x), then show that f(1/x) .f(1/y) - (1/2)[ f(x/ y) + f(x y) ] = 0.

Answer 1

I am very junior l don't know about gp sorry

Answer 2

Solution:

$f(x) = \cos( log(x) ) \\ \\ f( \frac{1}{x} ) = cos \: ( log\frac{1}{x} ) \\ \\ = cos(log \: 1 - log(x)) \\ \\ = cos \: (0 - log(x)) \\ \\ = cos \: ( - log \: x) \\ \\ f( \frac{1}{x}) = cos \: (log \: x) \\\\since\:cos(-\theta)=cos(\theta) \\by \: \: the \: \: same \: \: way \\ \\ f( \frac{1}{y} ) = cos \: ( log\frac{1}{y} ) = cos \: (log \: y)$
$f( \frac{1}{x} ).f( \frac{1}{y}) = cos \: log(x) .cos \: log(y) ....eq1\\ \\ f( \frac{x}{y} ) = cos \: log( \frac{x}{y} ) \\ \\ = cos \: (log \: x - log \: y) \\ \\ f(xy) = cos \: log(xy)$
$f( \frac{x}{y} ) + f(xy) = cos \: (log \: x - log \: y) + cos \: (log \: x + log \: y) \\ \\ as \: cos \: (A - B) + cos(A + B) = 2cos \: A \: cos \: B \\ \\ = 2 \: cos \: (log \: x )\: cos \: (log \: y)...eq2$
now put both eq1 and eq2 in the

$f( \frac{1}{x} ).f( \frac{1}{y}) - \frac{1}{2} (f (\frac{x}{y}) .f(xy)) \\ \\ = cos \: (log \: x)cos \: (log \: y) - \frac{1}{2} (2 \: cos(log \: x)cos(log \: y)) \\ \\ = cos \: (log \: x)cos \: (log \: y) -cos \: (log \: x)cos \: (log \: y) - \\ \\ = 0$
hence proved