已知tan2a=-3/4,π/4<a<π/2,
(1)tana的值;
tan2a=2tana/[1-(tana)^2]
2tana/[1-(tana)^2]=-3/4→
8tana=-3*[1-(tana)^2]→
8tana=-3+3(tana)^2→
3(tana)^2-8tana-3=0→
(tana-3)(3tana+1)=0
π/4<a<π/2→tana>0,→3tana+1>0
∴tana-3=0,
tana=3
(2)[2cos^(a/2)-sina-1]/[√2sin(a+π/4)]=
[2cos^(a/2)-1-sina]/[√2sin(a+π/4)]=
[cosa-sina]/[√2sin(a+π/4)]=
√2[(√2/2cosa-√2/2sina]/[√2sin(a+π/4)]=
√2[sin(π/4-a]/[√2sin(a+π/4)]=
sin(π/4-a]/sin(π/4+a)=
sin(π/4-a]/cos[π/2-(π/4+a)]=
sin(π/4-a]/cos(π/4-a)=
tan(π/4-a)=
(tanπ/4-tana)/(1+tanπ/4*tana)
(1-3)/(1+1*3)=
-2/4=
-1/2
