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p与q的夹角为π/31.求角β的大小2.已知tanC=根号3/2,求(sin2AcosA-sinA)/(sin2Acos2A)
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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p与q的夹角为π/31.求角β的大小2.已知tanC=根号3/2,求(sin2AcosA-sinA)/(sin2Acos2A)](/uploads/image/z/3755381-5-1.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%29%2Cq%3D%28cosB%2F2%2C-sinB%2F2%29%2C%E4%B8%94p%E4%B8%8Eq%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA%CF%80%2F31.%E6%B1%82%E8%A7%92%CE%B2%E7%9A%84%E5%A4%A7%E5%B0%8F2.%E5%B7%B2%E7%9F%A5tanC%3D%E6%A0%B9%E5%8F%B73%2F2%2C%E6%B1%82%EF%BC%88sin2AcosA-sinA%EF%BC%89%2F%EF%BC%88sin2Acos2A%EF%BC%89)
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p与q的夹角为π/31.求角β的大小2.已知tanC=根号3/2,求(sin2AcosA-sinA)/(sin2Acos2A)
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p与q的夹角为π/3
1.求角β的大小
2.已知tanC=根号3/2,求(sin2AcosA-sinA)/(sin2Acos2A)
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p与q的夹角为π/31.求角β的大小2.已知tanC=根号3/2,求(sin2AcosA-sinA)/(sin2Acos2A)
1.(cosB/2的表达不是很清楚,我理解的是cos(B/2),如果是(cosB)/2的话,LZ再追问好了)
p与q的夹角为π/3,
所以cosπ/3=pq/(|p|*|q|)=1/2
|p|^2=(cosB/2)^2+(sinB/2)^2=1
|q|^2=(cosB/2)^2+(-sinB/2)^2=1
所以pq=(cosB/2)^2-(sinB/2)^2=(cosπ/3)*(|p|*|q|)=1/2
所以(cosB/2)^2-(sinB/2)^2=cosB=1/2
所以B=π/3
2.
化简:(sin2AcosA-sinA)/(sin2Acos2A)
= [sinA*(2*cosA*cosA - 1)]/[2*sinA*cosA*cos2A] (消去sinA)
= [2(cosA)^2 - 1]/[2cosAcos2A]
= cos2A/[2cosA·cos2A]
= 1/(2cosA)
只需要求出cosA即可
tanC =根号3/2 ,所以sinC=根号3/2cosC
所以(sinC)^2=3/4(cosC)^2
又因为(sinC)^2+(cosC)^2=1
所以(sinC)^2+(cosC)^2=(sinC)^2+4/3(sinC)^2=7/3(sinC)^2=1
所以(sinC)^2=3/7,(cosC)^2=4/7
所以C是锐角,cosC>0
所以sinC= (根号21)/7 ,cosC = (2根号7)/7 ,
又cosB = 1/2 ,sinB = 根号3/2 ,
所以cosA = cos[π -(B+C)]= -cos(B+C)= sinB·sinC - cosB·cosC =1/(2根号7),
所以1/(2cosA) = 根号7 ,
也就是 (sin2A·cosA-sinA)/(sin2A·cos2A) = 根号7