已知向量a=(sinθ,1)b=(cosθ,2),满足a平行b，其中θ∈（0，π⼀2）若sin（θ+ψ）=1⼀3，求sinΨ

向量a=(sin$,1)b=(cos$,2),向量a//b,则有sin$*2-cos$=0
即有tan$=1/2,得到sin$=1/根号5,cos$=2/根号5
又有sin($+@)=1/3,故有cos($+@)=(+/-)根号(1-1/9)=(+/-)2根号2/3

sin@=sin($+@-$)=sin($+@)cos$-cos($+@)sin$
=1/3*2根号5/5(-/+)2根号2/3*根号5/5
=(2根号5(-/+)2根号10)/15

2a=(2sinθ,2)
b=(cosθ,2)
a//b
2sinθ=cosθ
tanθ=1/2=y/x,y取1，x取2，==>r=√5
sinθ=1/√5,cosθ=2/√5
sin(θ+ψ)=1/3,cos(θ+ψ)=±2√2/3
sinψ=[sin(θ+ψ)-θ]=sin(θ+ψ)cosθ-cos(θ+ψ)sinθ
1）当cos(θ+ψ)=2√2/3
sinψ=sin(θ+ψ)cosθ-cos(θ+ψ)sinθ=(1/3)(2/√5)-(2√2/3)(1/√5)=(2-2√2)/3√5
2）当cos(θ+ψ)=－2√2/3
sinψ=sin(θ+ψ)cosθ-cos(θ+ψ)sinθ=(1/3)(2/√5)+(2√2/3)(1/√5)=(2+2√2)/3√5