Question

在△ABC中，2(sinB)^2+3(sinC)^2=2sinAsinBsinC+(sinA)^2,求tanA

Answer 1

3sinB^2+3sinC^2-2sinBsinC=3sinA^2
由正弦定理得 a/sinA=b/sinB=c/sinC
sinB=b/a ·sinA sinC=c/a ·sinA
化简得 3sinB^2+3sinC^2+3sinA^2=2sinBsinC
3﹙b2/a2﹚sin2A+3﹙c2/a2﹚sin2A+3sin2A=2· ﹙b/a﹚sinA·﹙c/a﹚sinA
同时除以 sin2A/a2 3b2+3c2-3a2=2bc
3·﹙b2+c2-a2﹚/2bc =1
cosA=1/3
∵sin2A+cos2A=1
∴sin2A+﹙1/3﹚2=1
∴sinA= 2√2 / 3
向量AB*向量AC=bc·cosA
有3b^2+3c^2-2bc=3a^2,得3(b^2+c^2)-2bc=9,又b^2+c^2大于等于2bc,所以上式可得4bc小于等于9,故向量AB*向量AC=bc·cosA=(1/3)bc的最大值为3/4.