Question

在三角形ABC中，ABC所对边分别为abc，面积为S，已知tanC=2,2(sin^2B-sin^2A)=sin^2

在三角形ABC中，ABC所对边分别为abc，面积为S，已知tanC=2,2(sin^2B-sin^2A)=sin^2C
（1）求角A大小
（2）若S=15，求b值

Answer 1

解：sin^2B+sin^2C-sinBsinC=sin^2A，由正弦定理得b²+c²-bc=a²b²+c²-a²=bc=2bccosA解得cosA=1/2A=π/3正弦定理a/2R=sinAR=1∴BC=√3方法一：向量AB•向量AC=AB·AC·cosAcosA为定值当AB=AC时值最大，为[√(AB²+AC²)]/2向量AB•向量AC=AB·AC·cosA=√3×√3×1/2=3/2方法二：向量AB•向量AC=AB·AC·cosA=2RsinC·2RsinB·cosA=4R²sinC·sinB·cosA4R²是定值，cosA是定值。∴求向量AB•向量AC的最大值就是求sinC·sinB的最大值。∴sinC·sinB=sinC·sin(2π/3-C)=sinC(√3/2cosC+1/2sinC)=√3/2sinCcosC+1/2sin²C=√3/4sin2C+1/2[(1-cos2C)/2]=√3/4sin2C+1/4-1/4cos2C=√3/4sin2C-1/4cos2C+1/4=1/2(√3/2sin2C-1/2cos2C)+1/4=1/2(cosπ/6sin2C-sinπ/6cos2C)+1/4=1/2sin(2C-π/6)+1/4当C=π/3时，sin(2C-π/6)最大，1/2sin(2C-π/6)+1/4最大，为3/4∴向量AB•向量AC=AB·AC·cosA=2RsinC·2RsinB·cosA=4R²sinC·sinB·cosA=4×3/4×1/2=3/2