设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=(1,0),α∈(0,π),β∈(π,2π),设a与c的夹角为θ1,b与c的夹角为θ2,且θ1-θ2=π/6.求sin[(α-β)/8]的值.
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![设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=(1,0),α∈(0,π),β∈(π,2π),设a与c的夹角为θ1,b与c的夹角为θ2,且θ1-θ2=π/6.求sin[(α-β)/8]的值.](/uploads/image/z/6084483-51-3.jpg?t=%E8%AE%BEa%3D%EF%BC%881%EF%BC%8Bcos%CE%B1%2Csin%CE%B1%EF%BC%89%2Cb%3D%EF%BC%881%EF%BC%8Dcos%CE%B2%2Csin%CE%B2%EF%BC%89%2Cc%3D%EF%BC%881%2C0%EF%BC%89%2C%CE%B1%E2%88%88%EF%BC%880%2C%CF%80%EF%BC%89%2C%CE%B2%E2%88%88%EF%BC%88%CF%80%2C2%CF%80%EF%BC%89%2C%E8%AE%BEa%E4%B8%8Ec%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA%CE%B81%2Cb%E4%B8%8Ec%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA%CE%B82%2C%E4%B8%94%CE%B81%EF%BC%8D%CE%B82%3D%CF%80%2F6.%E6%B1%82sin%5B%EF%BC%88%CE%B1%EF%BC%8D%CE%B2%EF%BC%89%2F8%5D%E7%9A%84%E5%80%BC.)
设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=(1,0),α∈(0,π),β∈(π,2π),设a与c的夹角为θ1,b与c的夹角为θ2,且θ1-θ2=π/6.求sin[(α-β)/8]的值.
设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=(1,0),α∈(0,π),β∈(π,2π),设a与c的夹角为θ1,b与c的夹角为θ2,且θ1-θ2=π/6.求sin[(α-β)/8]的值.
设a=(1+cosα,sinα),b=(1-cosβ,sinβ),c=(1,0),α∈(0,π),β∈(π,2π),设a与c的夹角为θ1,b与c的夹角为θ2,且θ1-θ2=π/6.求sin[(α-β)/8]的值.
解,由题意可知,
cosθ1=a*c/(|a|*|c|=√{(1+cosa)/2}
cosθ2=b*c/(|b|*|c|)=√{(1-cosb)/2},所以,θ1,θ2都在(0,π/2)之间
由于,α∈(0,π),那么α/2∈(0,π/2),1+cosa=2cos²(a/2),√(1+cosa)/2=cos(a/2)
也即是,θ1=a/2
由于,β∈(π,2π),那么β/2∈(π/2,π),1-cosβ=2sin²(β/2),√(1-cosβ)/2=sin(β/2),又由于sinβ/2=cos(π/2-β/2)=cos(β/2-π/2),因此θ2=β/2-π/2
θ1-θ2=π/6,故,a/2-β/2=-π/3
那么,a/8-β/8=-π/12
又由于sin(π/6)=2*sin(π/12)*cos(π/12)=1/2,sin²(π/12)+cos²(π/12)=1
可以解出,sin(π/12)=(√6-√2)/4,
所以,sin(-π/12)=(√2-√6)/4
故,sin[(α-β)/8]=sin(-π/12)=(√2-√6)/4