在三角形ABC中,AB=2根号5,AC=3,sinC=2sinA,求三角形ABC面积

为方便起见,设c=AB ,b=AC ,a=BC∵ AB=2根号5,AC=3,∴ c=2√5, b=3∵ sinC=2sinA利用正弦定理a/sinA=b/sinB=c/sinC∴ c=2a∴ a=√5∴ cosA=(b²+c²-a²)/(2bc)=(9+20-5)/(2*3*2√5)=2/√5∴ sinA=1/√5=...还有个问求cos2A＋4分之派sinA=1/√5，cosA=2/√5sin2A=2*sinA*cosA=4/5cos2A=1-2sin²A=3/5cos(2A+π/4)=cos2A*cos(π/4)-sin2A*sin(π/4)=(3/5)*(√2/2)-(4/5)*(√2/2)=-√2/10