Question

已知sin(π/4+2a)sin(π/4-2a)=1/6....

已知sin(π/4+2a)sin(π/4-2a)=1/6,a∈（π/4，π/2）
（1）求cos4a的值
（2）求2根3sin^2a+tana-cota-根3的值

Answer 1

sin(π/4+2A)sin(π/4-2A)

=[cos4A-cos(π/2)]/2

=(cos4A)/2=1/6

故cos4A=1/3,

而A∈（π/4，π/2),2A∈（π/2，π),4A∈（π，2π)

cos4A=2cos^2 2A-1

1/3=2cos^2 2A-1

2cos^2 2A=4/3

cos^2 2A=2/3

cos2A=-√6/3

sin^2 2A=1/3

sin2A=√3/3

2√3sin^2 A+tanA-cotA-√3

=√3(2sin^2 A-1)+(sin^2 A-cos^2 A)/sinAcosA

=-√3cos2A+(sin^2 A-cos^2 A)/sinAcosA

=-√3cos2A-2cos2A/sin2A

=√3cos2A(1-2/sin2A)

=√3(-√6/3)[1-2/(√3/3)]

=-√2[1-2√3]

=2√6-2