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设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=（1，0），α∈(0,π),β∈(π,2π),a与c的

设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=（1，0），α∈(0,π),β∈(π,2π),a与c的夹角为θ₁,b与c的夹角为θ₂,且θ₁-θ₂=,求sin的值.

答案

解析：∵a

=(2 cos²,2sincos)

=2 cos(cos,sin),

又∵cosθ₁=,∴cosθ₁=cos［α∈(0,π),∈(0,］.

∴θ₁=.

又∵b

=(2sin²,2 sincos)=2sin(sin,cos),

∴cosθ₂=sin［β∈(π,2π),-∈(0,］.

∴θ₂=-.

又θ₁-θ₂=-+==-,∴sin=sin(-)=-.

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设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=（1，0），α∈(0,π),β∈(π,2π),a与c的

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