f(x)=√2（sinxcosπ/4+cosxsinπ/4)=)=√2sin(x+π/4)解法一：x∈(-π/2,π/2)则(x+π/4)∈(-π/4,3π/4)令Y=x+π/4,则g(Y)=√2sinY又sinY,在Y∈[-π/4,π/2]单调递增,在Y∈[π/2,3π/4]单调递减,故f(x)max=g(Y)max...