Question

急>< TAT已知向量a=（sinx,1),向量b=（cosx,-1/2) （1）当a⊥b,求丨a+b丨的值 (2)求函

Answer 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

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]