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),且向量p与向量q的夹角为π/3.(1)求角B的大小(2)已知tanC=根号3/2,求(sin2AcosA-sinA)/sin2Acos2A的值
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![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),且向量p与向量q的夹角为π/3.(1)求角B的大小(2)已知tanC=根号3/2,求(sin2AcosA-sinA)/sin2Acos2A的值](/uploads/image/z/2533992-24-2.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CA%2CB%2CC%E4%B8%BA%E5%AE%83%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E8%AE%BE%E5%90%91%E9%87%8Fp%3D%28cosB%2F2%2CsinB%2F2%EF%BC%89%2Cq%3D%28cosB%2F2%2C-sinB%2F2%29%2C%E4%B8%94%E5%90%91%E9%87%8Fp%E4%B8%8E%E5%90%91%E9%87%8Fq%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA%CF%80%2F3.%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92B%E7%9A%84%E5%A4%A7%E5%B0%8F%EF%BC%882%29%E5%B7%B2%E7%9F%A5tanC%3D%E6%A0%B9%E5%8F%B73%2F2%2C%E6%B1%82%EF%BC%88sin2AcosA-sinA%29%2Fsin2Acos2A%E7%9A%84%E5%80%BC)
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),且向量p与向量q的夹角为π/3.(1)求角B的大小(2)已知tanC=根号3/2,求(sin2AcosA-sinA)/sin2Acos2A的值
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),且向量p与向量q的夹角为π/3.
(1)求角B的大小
(2)已知tanC=根号3/2,求(sin2AcosA-sinA)/sin2Acos2A的值
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),且向量p与向量q的夹角为π/3.(1)求角B的大小(2)已知tanC=根号3/2,求(sin2AcosA-sinA)/sin2Acos2A的值
(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