在△ABC中,A,B,C为它的三个内角,设向量p=(cosB/2,sinB/2）,q=(cosB/2,-sinB/2),

(1) p=(cosB/2,sinB/2）,q=(cosB/2,-sinB/2),
则IpI=1 IqI=1
p*q=cos²B/2-sin²B/2=cosB
又p*q=IpI*IqIcos(π/3)=1/2
所以cosB=1/2
故B=π/3
(2) 已知tanC=根号3/2
则tanA=-tan(B+C)=-(tanB+tanC)/(1-tanBtanC)=-(√3+√3/2)/(1-√3*√3/2)=3√3
secA=√(1+tan²A)=2√7
（sin2AcosA-sinA)/sin2Acos2A
=sinA(2cos²A-1)/2sinAcosAcos2A
=cos2A/2cosAcos2A
=1/(2cosA)
=secA/2
=2√7/2
=√7
希望能帮到你O(∩_∩)O