If for all real values of `ua n dv ,2f(u)cosv=f(u+v)+f(u-v),` prove that for all real values of `x ,` `f(x)+f(-x)=2acosxdot` `f(pi-x)+f(-x)=0` `f(pi-x)+f(x)=2bsinxdot` Deduce that `f(x)=acosx+bsinx ,w h e r ea ,b` are arbitrary constants.

Explanation

Given `2f(u) cos v =f(u+v)+f(u-v) " (1)" `
Putting `u=0` and `v=x` in (1) we get
`f(x) +f(-x)=2f(0)cos x+2a cosx " (2) " `
a is an arbitrary contant.
Now, putting `u=(pi)/(2) -x " and " v=(pi)/(2)` in (1), we get
`f(pi- x) +f(-x)=0 " (3) " `
Again, putting `u=pi//2` and `v=pi//2-x ` in (1), we get
`f(pi-x)+f(x) =2f(pi//2)sin x=2b sinx " (4) " `
b is an arbitrary constant.
Adding (2) and (4), we get
`2f(x) +f(pi-x)+f(-x) = 2a cosx +2b sinx`
or ` 2f(x) +0=2a cosx +2b sinx " [From (3)] " `
` :. f(x) =a cosx +b sinx`