已知向量a=(sinx,1),b=(cosx,-1/2)

1.a⊥b时,a,b构成直角三角形,
|a+b|^2=|a|^2+|b|^2=(sinx^2+1)+(cosx^2+1/4)=9/4
=>|a+b|=3/2
2.
f(x)
=a(a-b)
=a^2-a.b
=sinx^2+1-(sinxcosx-1/2)
=(1-cos2x)/2-sin2x/2+3/2
=2-(sin2x+cos2x)/2
=2-√2/2sin(2x+π/4)
∈[2-√2/2,2+√2/2]