又f(π/8)=-1,f(3π/8)=√2-1,f(3π/4)=√2sin(3π/2 "-" π/4)-1=-√2cosπ/4-1=-2,

　　故函数f(x)在区间[π/8 "," 3π/4]上的最大值为√2-1,最小值为-2.

B组　能力提升

1.(2cos^2 x"-" 1)/(2tan(π/4 "-" x)sin^2 (π/4+x) )可化简为(　　)

A.1 B.-1 C.cos x D.-sin x

解析原式=cos2x/(2tan(π/4 "-" x)cos^2 (π/4 "-" x) )

　　=cos2x/(2"·" sin(π/4 "-" x)/cos(π/4 "-" x) cos^2 (π/4 "-" x) )

　　=cos2x/2sin(π/4 "-" x)cos(π/4 "-" x)

　　=cos2x/sin(π/2 "-" 2x) =1.

答案A

2.若cos θ=-4/5,θ是第三象限的角,则(1"-" tan θ/2)/(1+tan θ/2)=(　　)

A.1/2 B.-1/2 C.3/5 D.-2

解析(1"-" tan θ/2)/(1+tan θ/2)=(cos θ/2 "-" sin θ/2)/(cos θ/2+sin θ/2)

　　=(cos θ/2 "-" sin θ/2)(cos θ/2+sin θ/2)/(cos θ/2+sin θ/2)^2 =cosθ/(1+sinθ),

　　因为cos θ=-4/5,且θ是第三象限的角,

　　所以sin θ=-3/5,故(1"-" tan θ/2)/(1+tan θ/2)=("-" 4/5)/(1"-" 3/5)=-2.

答案D

3.若cos2α/sin(α"-" π/4) =-√2/2,则cos α+sin α的值为　　　.

解析∵cos2α/sin(α"-" π/4) =(cos^2 α"-" sin^2 α)/(sinαcos π/4 "-" cosαsin π/4)

　　=(√2 "(" cosα+sinα")(" cosα"-" sinα")" )/(sinα"-" cosα)

　　=-√2(cos α+sin α)=-√2/2,

　　∴cos α+sin α=1/2.

答案1/2

4.已知角α,β为锐角,且1-cos 2α=sin αcos α,tan(β-α)=1/3,则β=　　　　　.

解析由1-cos 2α=sin αcos α,得1-(1-2sin2α)

　　=sin αcos α,

　　即2sin2α=sin αcos α.

　　∵α为锐角,∴sin α≠0,

　　∴2sin α=cos α,即tan α=1/2.

　　(方法一)由tan(β-α)=(tanβ"-" tanα)/(1+tanβtanα)=(tanβ"-" 1/2)/(1+1/2 tanβ)

=1/3,得tan β=1.∵β为锐角,∴β=π/4.