若sinα=(m-3)/m+5,cos=(4-2m）/m+5,π/2＜α＜π，则m=
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解决时间 2021-01-30 11:28
- 提问者网友：最爱你的唇
- 2021-01-30 02:14
若sinα=(m-3)/m+5,cos=(4-2m）/m+5,π/2＜α＜π，则m=
最佳答案
- 五星知识达人网友：胯下狙击手
- 2021-01-30 02:57
π/2<а<π,
(m-3)/(m+5)=sin(a)>0, (m-3)(m+5)>0, m<-5或m>3
(4-2m)/(m+5)=cos(a)<0, (4-2m)(m+5)<0, (m-2)(m+5)>0, m>2或m<-5
m>3 或m<-5
1=[sin(a)]^2+[cos(a)]^2=[(m-3)/(m+5)]^2+[(4-2m)/(m+5)]^2
(m+5)^2=(m-3)^2 + (2m-4)^2
m^2+10m+25=m^2-6m+9+4m^2-16m+16
0=4m^2-32m=4m(m-8)
m=0(舍去)m=8
∴m=8
(m-3)/(m+5)=sin(a)>0, (m-3)(m+5)>0, m<-5或m>3
(4-2m)/(m+5)=cos(a)<0, (4-2m)(m+5)<0, (m-2)(m+5)>0, m>2或m<-5
m>3 或m<-5
1=[sin(a)]^2+[cos(a)]^2=[(m-3)/(m+5)]^2+[(4-2m)/(m+5)]^2
(m+5)^2=(m-3)^2 + (2m-4)^2
m^2+10m+25=m^2-6m+9+4m^2-16m+16
0=4m^2-32m=4m(m-8)
m=0(舍去)m=8
∴m=8
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