a=(cos(3x/2),sin(3x/2))
b=(cos(x/2),-sin(x/2))
因此,
a·b
=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(3x/2+x/2)
=cos(2x)
|a+b|
=√(a+b)^2
=√(a^2+b^2+2ab)
=√(1+1+2cos(2x))
=√2√(1+cos(2x))
=√2√(1+2cos^2x-1)
=2|cosx|
因?yàn)閤∈[0,π/2],cosx∈[0,1]
=2cosx
f(x)
=a·b-4m|a+b|+1
=cos(2x)-8mcosx+1
=2cos^2x-8mcosx
為求簡化表達(dá)式,換元t=cosx∈[0,1]
f(x)=h(t)=2t^2-8mt
對(duì)對(duì)稱軸t=2m分類討論：
1.2m≥1：h(t)單調(diào)遞減,g(m)=f(x)min=h(1)=2-8m
2.2m≤0：h(t)單調(diào)遞增,g(m)=f(x)min=h(0)=0
3.0