b^2=ac,cos B=3/4 求cot A+cot C2.设向量BA*向量BC=3/2,求a+c的值
问题描述：
b^2=ac,cos B=3/4 求cot A+cot C
2.
设向量BA*向量BC=3/2,求a+c的值
答
sinB=根号7/4
cot A+cot C=sin(A+C)/(sinAsinC)=sinB/(sinAsinC)
=[(sinB)^2/(sinAsinC)]/sinB=[b^2/(ac)]/sinB=4/根号7
2.向量BA*向量BC=accosB=3/2
ac=b^2=2
cosB=(a^2+c^2-b^2)/(2ac)=(a^2+c^2-2)/(2ac)=3/4
求出a^2+c^2
（a+c）^2=a^2+c^2+2ac
答
(1)cotA+cotC=4√7/7.
由cosB=3/4得sinB=√(1-(cosB)^2)=√7/4
cotA+cotC=cosA/sinA+cosC/sinC
=sin(A+C)/(sinAsinC)=sin(180-B)/(sinAsinC)=sinB/(sinAsinC)
由正弦定理sinA=asinB/b,sinC=csinB/b,和b^2=ac,得
cotA+cotC=sinB/(sinAsinC)=b^2/(acsinB)=1/sinB=4√7/7
(2)a+c=3
由向量BA*向量BC=3/2,
accosB=3/2,ac=(3/2)/cosB=2,b^2=ac=2
由余弦定理得
2=b^2=a^2+c^2-2accosB=a^2+c^2-3,
a^2+c^2=5,(a+c)^2=a^2+c^2+2ac=9,故得a+c=3,或a+c=-3(不合题意舍去).