{cos[派/2)+A]}的平方+cosA=5/4
即(sinA)^2+cosA=5/4
即(cosA)^2-cosA+¼=0
(cosA-1/2)^2=0
cosA=1/2
三角形ABC,所以A<180°
则sinA=√3/2 .A=60°
b+c=√3a
由正弦定理：
即sinB+sinC=√3sinA
即
2sin(B+C)/2cos(B-C)/2=√3*√3/2
而A+B+C=180°
则(B+C)/2=90°-A/2
则sin(B+C)/2=sin(90°-A/2)=cosA/2=cos30°=√3/2
则可知,
cos(B-C)/2=√3/2
则cos(B-C)=2（cos(B-C)/2）^2-1=1/2
2.
Z=sin(B+C)-icos(B-C)
=sinA-icos(B-C)
=√3/2-1/2i
Z^2+（1/Z^2）-1
=(Z+1/Z)^2-3
=(√3/2-1/2i+√3/2+1/2i)^2-3
=3-3
=0