题目

已知a∈（π4,π2),cos(π4+2a)cos(π4-2a)=14.(1)求a的值.（2）求2sin²a+tana-1 ana-1的值

提问时间：2020-12-22

答案

cos(π/4+2a)cos(π/4-2a)=1/4
cos(π/4+2a)cos(π/2-π/4-2a)=1/4
cos(π/4+2a)cos[π/2-(π/4+2a)]=1/4
cos(π/4+2a)sin(π/4+2a)=1/4
1/2sin(π/2+4a)=1/4
sin(π/2+4a)=1/2
cos4a=1/2
a∈（π/4,π/2)
4a∈（π,2π)
cos4a=1/2
4a∈（3π/2,2π)
4a=5π/3
a=5π/12
4a∈（3π/2,2π)
2a∈（3π/4,π)
cos4a=1/2
2cos²2a-1=1/2
cos²2a=3/4
cos2a=-√3/2
2cos²a-1=-√3/2
cos²a=(2-√3)/4
sin²a=1-cos²a
=1-(2-√3)/4
=(2+√3)/4
=2*(4+2√3)/16
sina=(√6+√2)/4
cos²a=(2-√3)/4
cos²a=2(4-2√3)/16
cosa=(√6-√2)/4
tana=sina/cosa
=[(√6+√2)/4]/[(√6-√2)/4]
=(√6+√2)/(√6-√2)
=(√6+√2)²/(√6-√2)(√6+√2)
=(8+4√3)/4
=2+√3
(2sin²a+tana-1)/(tana-1)
=2sin²a/(tana-1)+(tana-1)/(tana-1)
=2sin²a/(tana-1)+1
=[2*(2+√3)/4]/(2+√3-1)+1
=[(2+√3)/2]/(√3+1)+1
=[(2+√3)(√3-1)/2]/(√3+1)(√3-1)+1
=[(2+√3)(√3-1)/2]/2+1
=[(√3+2)(√3-1)/2]/2+1
=[(3+√3-2)/2]/2+1
=(1+√3)/4+1
=(5+√3)/4