Question

y=2sinx(sinx+cos),x∈[0,π/2]最大值为?3Q

Answer 1

y=2sinx^2+2sinxcosx =1-cos2x+sin2x =√2(1/2sin2x-√2/2cos2x)+1 =√2cosπ/3sin2x-sinπ/3cos2x)+1 =√2sin(2x-π/3)+1 x∈[0,π/2] -π/3≤2x-π/3≤2π/3 当2x-π/3=π/2即 ,当x=5π/12取得最大值为√2+1 当2x-π/3= -π/3即 ,当x=0最小值为√6/2+1