Question

数学求解！！！！

Answer 1

【参考答案】

(1)sinC=2sinA即c=2a
∵a=√5，b=3
∴c=2√5
即BC=2√5

（2）由余弦定理得
cosA=(9+20-5)/(2×3×2√5）=2/√5
∴sinA=1/√5，sin2A=2×(1/√5)×(2/√5)=4/5
cos2A=1-2sin²A=1-2×(1/5)=3/5
∴sin(2A- π/4)
=(4/5)×(√2/2)-(3/5)×(√2/2)
=√2/10

Answer 2

1、∵sinC=2sinA
∴AB/2R=2×BC/2R
AB=2×BC=2√5
2、
sin(2A-π/4)=sin2Acos(π/4)-cos2Asin(π/4)
=[(√2)/2](sin2A-cos2A)
利用余弦定理求角A:
cosA=(AB²+AC²-BC²)/2AB*AC
=[(2√5)²+3²-(√5)²]/2×(2√5)×3
=(20+9-5)/12(√5）
故，cosA=(2√5)/5
sinA=√[1-cos²A]=(√5)/5
sin(2A-π/4)=[(√2)/2][2sinAcosA-(2cos²A-1)]
=[(√2)/2]{2×(√5/5)×(2√5/5)-[2×(2√5/5)²-1]}
整理后得：
sin(2A-π/4)=(√2)/10

Answer 3

AB=2√5