Question

已知向量a=(sinФ，cosФ-2sinФ),b=(1,2) 若向量|a|=|b|,0＜Ф＜π，求Ф

Answer 1

依题意得：(sinФ)^2+(cosФ-2sinФ)^2=1^2+2^2
即(sinФ)^2+(cosФ)^2+4(sinФ)^2-4sinФcosФ=5
即1+4(sinФ)^2-2sin2Ф=5，而2(sinФ)^2=1-cos2Ф
所以1+2(1-cos2Ф)-2sin2Ф=5
所以sin2Ф+cos2Ф=-1
即√2sin(2Ф+π/4)=-1
所以sin(2Ф+π/4)=-√2/2
而0＜Ф＜π，则π/4<2Ф+π/4<9π/4
所以2Ф+π/4=5π/4或7π/4
则Ф=π/2，或Ф=3π/4

Answer 2

(sinФ)^2+(cosФ)^2-4cosФ sinФ+4(cosФ)^2=1+2^2
1-2sin2Ф+2[2(cosФ)^2-1] +2 = 5
cos2Ф-sin2Ф=1
因为(cos2Ф)^2+(sin2Ф)^2=1
所以cos2Ф=0或1
所以2Ф=π/2 或3π/2 或0
Ф=π/4 或3π/4 或0

Answer 3

|b|=√5
|a|=√(sin^2Ф+(cosФ-2sinФ)^2)
=√(sin^2Ф+cos^2Ф-4sinФcosФ+4sin^2Ф)
=√(1-2sin2Ф+2(1-cos2Ф))
=√(3-2(sin2Ф+cos2Ф))
=√(3-2√2sin(2Ф+π/4))=√5
2√2sin(2Ф+π/4)=2
sin(2Ф+π/4)=√2/2 0＜Ф＜π π/4<2Ф+π/4< 9π/4
2Ф+π/4=3π/4
Ф=π/4