It is known that x 1, x2 is the function f (x) = sin (wx+q) (w >; 0，0 & lt； Q < π) and |x 1-x2|=π point (π,-1) is in the function.

The image of f(x): 1. Expression 2. If a belongs to (0, π/2) and f(2a)=3/5, find the value of 1/(sina+cosa).

Solution: ωx? +q=0, then x? =-q/ω...........( 1)

ωx？ +q=π, then x? =(π-q)/ω............(2)

︱x？ -x？ = (π-q)/ω+q/ω = π/ω = π, so ω= 1,

Then f(x)=sin(x+q), f(π)=sin(π+q)=-sinq=- 1, so q=π/2.

The expression ∴f(x) is f(x)=sin(x+π/2)=cosx.

f(2α)=cos(2α)=2cos？ α- 1=3/5，cos？ α=4/5，cosα= 2/√5； Sin? α= 1-4/5= 1/5，sinα= 1/√5

∴ 1/(sina+cosa)= 1/( 1/√5+2/√5)=(√5)/3.