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

1个回答

  • 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