(1)1°当r=0时,显然0≥-
3
8
2°当r≠0时,设φ(x)=rcosx+r2cos2x=r2(2cos2x-1)+rcosx
=2r2(cosx+
1
4r)2-
1
8-r2≥-
1
8-r2≥-
1
8-(
1
2)2=-
3
8.(|r|≤
1
2)
(2)当|r|≤
1
2时,∀x∈R,∀n∈N*(n≥2),f2n+1=
1
2+rcosx+r2cos2x+r3cos4x+r4cos8x++r2n-1cos22(n-1)x+r2ncos22n-1x
=
1
2+φ(x)+r2φ(4x)++r2(n-1)•φ(4n-1x)≥
1
2-
3
8(1+r2++r2(n-1))≥
1
2-
3
8(1+
1
4++
1
4n-1)
=
1
2-
3
8•
1-
1
4n
1-
1
4=
1
2•4n=
1
22n+1>0.