Question

已知向量a=（sinα，cosα-2sinα），b=（1，2），若|a|=|b|，0<α<π，求α

Answer 1

|a|=|b| => sin^2(α)+(cosα-2sinα)^2=1+2^2=5
=>sin^2(α)+ cos^2(α)-4sinαcosα+4sin^2(α)=5
=>1-2sin(2α)+2(1-cos(2α))=5
=>-2sin(2α)-2cos(2α)=2
=> sin(2α)+cos(2α)=-1
√(2) sin(2α+π/4)=-1
=> sin(2α+π/4)=-√(2)/2
=> 2α+π/4=-π/4+2nπ 或者 -3π/4+2nπ
2α=-π/2+2nπ 或者-π+2nπ
α=-π/4+nπ或者-π/2+nπ
所以α=π/4+1π=3π/4

Answer 2

|a|=|b|
=>(sinα)^2+(cosα-2sinα)^2 = 5
5(sinα)^2 -4sinαcosα + (cosα)^2 =5
4(cosα)^2 + 4sinαcosα =0
cosα(cosα+sinα) =0
=> cosα+sinα =0
tanα = -1
α= 3π/4

追问

5(sinα)^2 -4sinαcosα + (cosα)^2 =5
4(cosα)^2 + 4sinαcosα =0
这两步变化错了吧
应该是4(sinα)^2 + 4sinαcosα =4吧

追答

5(sinα)^2 -4sinαcosα + (cosα)^2 =5
-4sinαcosα + (cosα)^2 = 5- 5(sinα)^2 = 5(cosα)^2
4(cosα)^2 + 4sinαcosα =0

Answer 3

90°或135°