Question

已知：sin(π/4+2A)·sin(π/4—2A）=1/4，A∈（π/4，π/2）。求2sin的平方A+tanA—cotA—1的值？

Answer 1

sin(π/4+2A)sin(π/4-2A)=[cos4A-cos(π/2)]/2=(cos4A)/2=1/4 故cos4A=1/2,而A∈（π/4，π/2),4A∈（π，2π) 故4A=5π/3,A=5π/12,2A=5π/6 sin2A=1/2,cos2A=-√3/2 则2sin^2 A+tanA-cotA-1 =-cos2A+(sin^2 A-cos^2 A)/sinAcosA =-cos2A-2cos2A/sin2A =5√3/2

麻烦采纳，谢谢!