Question

已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8)=-4/5,求|m+n|的值谢谢

Answer 1

θ∈(π,3π/2),且cos(θ/2+π/8)=-4/5,
所以sin(θ/2+π/8)=-3/5

因为m=(cosθ,-sinθ)和n＝(√2+sinθ,cosθ)
所以m+n=（√2+sinθ+ cosθ，-sinθ+cosθ）
所以
|m+n|
=√[(√2+sinθ+ cosθ)^2+(-sinθ+cosθ)^2]
=√[(2+(sinθ)^2+(cosθ)^2+2sinθcosθ+2√2( cosθ+sinθ)+ (sinθ)^2+(cosθ)^2-2sinθcosθ)]
=√[(2+1+2√2( cosθ+sinθ)+1]
=√[(4+2√2( cosθ+sinθ)]
=√[(4+2√2*√2( √2/2cosθ+√2/2sinθ)]
=√[(4+4( √2/2cosθ+√2/2sinθ)]
=√[4+4(sinπ/4cosθ+sinθcosπ/4)]
=√[4+4sin(θ+π/4)]
=2√[1+sin(θ+π/4)]
=2√[1+2sin(θ/2+π/8)cos(θ/2+π/8)]
=2√[1+2*(-3/5)*(-4/5)]
=2√[1+24/25]
=2√(49/25)
=2*7/5
=14/5