已知向量m=(cosθ,-sinθ),n=(根号2+sinθ,cosθ),θ∈(π,3π/2),且cos(θ/2+π/8

学习 时间:2026-04-04 19:37:58 阅读:733
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

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单薄的荷花

细腻的唇彩

2026-04-04 19:37:58

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

最新回答共有2条回答

  • 雪白的歌曲
    回复
    2026-04-04 19:37:58

    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

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