Question

在三角形ABC中，3sinB^2+7sinC^2=2sinAsinBsinC+2sinA^2,求sin(A+pi/4)

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.