Question

已知sin（π/4+2a）*sin（π/4-2a）=1/4，a属于（π/4，π/2），求2sin²

已知sin（π/4+2a）*sin（π/4-2a）=1/4，a属于（π/4，π/2），求2sin²a+tana-1/tana-1的值。

Answer 1

a∈(π/4,π/2)
2a∈(π/2,π)
cos2a0 sin2a0
4a∈(π,2π)
π/2+4a∈(3π/2,5π/2)
sin(π/4+2a)sin(π/4-2a) =1/4
2sin(π/4+2a)(-cos(π/4+2a))=1/2
-sin[2(π/4+2a)]=1/2
-sin(π/2+4a)]=1/2
cos4a=1/2由cos4a=2cos²2a-1=1/2 cos²2a=3/4
cos2a=-√3/2 (1)
又cos4a=1-2sin²2a=1/2 sin²2a=1/4
sin2a=1/2 (2)
2sin^2a+tana-1/tana-1
=-(1-2sin^2a)+sina/cosa-cosa/sina
=-cos2a+(sin^2a-cos^2a)/sinacosa
=-cos2a-2cos2a/2sinacosa
=-cos2a-2cos2a/sin2a
=√3/2-2(-√3/2)/(1/2)
=√3/2+2√3
=5√3/2