在△ABC ,BC=根号5,AC=3,sinC=2sinA (1)求AB 的值 (2)求sin(2A-派/4)的值

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/29 19:32:06
在△ABC,BC=根号5,AC=3,sinC=2sinA(1)求AB的值(2)求sin(2A-派/4)的值在△ABC,BC=根号5,AC=3,sinC=2sinA(1)求AB的值(2)求sin(2A-

在△ABC ,BC=根号5,AC=3,sinC=2sinA (1)求AB 的值 (2)求sin(2A-派/4)的值
在△ABC ,BC=根号5,AC=3,sinC=2sinA (1)求AB 的值 (2)求sin(2A-派/4)的值

在△ABC ,BC=根号5,AC=3,sinC=2sinA (1)求AB 的值 (2)求sin(2A-派/4)的值
sinC=2sinA
BC/sinA=AB/sinC
BC/AB=sinA/sinC=1/2
AB=2√5
cosA=(AB^2+AC^2-BC^2)/(2*AB*AC)
=(20+9-5)/(2*2√5*3)
=2√5/5
sinA=√5/5
sin2A=2sinAcosA=4/5
cos2A=2cos^2A-1=3/5
sin(2A-π/4)
=sin(2A)cos(π/4)-cos(2A)sin(π/4)
=4/5*√2/2-3/5*√2/2
=√2/10

(1)因BC对应于角A,AB对应于角C.应用正弦定理得:BC/sinA=AB/sinCAB=BC*sinC/sinA=BC*2sinA/sinA=2BC故,AB=2根号5.
(2) sin(2A-π/4)=sin2Acos(π/4)-cos2Asin(π/4)=[(根号2)/2](sin2A-cos2A)利用余弦定理求角A: cosA=(AB^+AC^2-BC^2)/2AB*AC=[(2...

全部展开

(1)因BC对应于角A,AB对应于角C.应用正弦定理得:BC/sinA=AB/sinCAB=BC*sinC/sinA=BC*2sinA/sinA=2BC故,AB=2根号5.
(2) sin(2A-π/4)=sin2Acos(π/4)-cos2Asin(π/4)=[(根号2)/2](sin2A-cos2A)利用余弦定理求角A: cosA=(AB^+AC^2-BC^2)/2AB*AC=[(2根号5)^2+3^2-(根号5)^2]/2*(2根号5)*3=(20+9-5)/12(根号5)故,cosA=(2根号5)/5sinA=根号[1-cos^2A]=(根号5)/5 sin(2A-π/4)
=[(根号2)/2][2sinAcosA-(2cos^2A-1)]=[(根号2)/2]{2*(根号5/5)*(2根号5/5)-[2*(2根号5/5)^2-1]}
整理后得:sin(2A-π/4)=(根号2)/10
【【不清楚,再问;满意, 请采纳!祝你好运开☆!!】】

收起