If `f: R->R` be given by `f(x)=s in^2x+s in^2(x+pi//3)+cosx\ cos(x+pi//3)` for all `x in R` , and `g: R->R` be such that `g(5//4)=1` , then prove that `gof: R->R` is a constant function.

Explanation

`f(x) = sin^(2) x + sin^(2)(x + (pi)/(3)) + cos x cos (x + (pi)/(3))`
`= sin^(2) x + sin^(2)(x + (pi)/(3)) + (1)/(2)[cos(2x + (pi)/(3)) + cos((pi)/(3))]`
`= (1-cos 2x + 1 - cos 2(x + (pi)/(3)) + cos.(pi)/(3)+cos(2x+(pi)/(3)))/(2)`
`= (5)/(4) + (1)/(2)[cos(2x + (pi)/(3)) - cos 2x-cos 2(x+(pi)/(3))]`
`= (5)/(4)+(1)/(2)[cos(2x + (pi)/()) - 2 cos (2x + (pi)/(3))cos ((pi)/(3))]`
`f(x) = (5)/(4)`
`g(f(x)) = g((5)/(4)) = 1` which is constant value hence (gof) : `R rarr R` is a constant function