Question

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),

Answer 1

(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