∴√(1"-" 2sin4cos4)=√("(" sin4"-" cos4")" ^2 )=|sin 4-cos 4|=cos 4-sin 4,故选C.

答案C

3.记cos(-80°)=k,那么tan 100°=(　　)

A.√(1"-" k^2 )/k B.-√(1"-" k^2 )/k

C.k/√(1"-" k^2 ) D.-k/√(1"-" k^2 )

解析因为sin 80°=√(1"-" cos^2 80"°" )=√(1"-" cos^2 "(-" 80"°)" )=√(1"-" k^2 ),所以tan 100°=-tan 80°=-sin80"°" /cos80"°" =-√(1"-" k^2 )/k.

答案B

4.已知sin α=(4"-" 2m)/(m+5),cos α=(m"-" 3)/(m+5),α是第四象限角,则tan α=　　　　　.

解析由sin2α+cos2α=1,知((4"-" 2m)/(m+5))^2+((m"-" 3)/(m+5))^2=1,解得m=8或m=0.

　　又α为第四象限角,则sin α<0,cos α>0,知m=8,则tan α=-12/5.

答案-12/5

5.已知tan α=2,则1/(1"-" sinαcosα)的值为　　　　　.

解析原式=(sin^2 α+cos^2 α)/(sin^2 α+cos^2 α"-" sinαcosα)=(tan^2 α+1)/(tan^2 α+1"-" tanα)=(4+1)/(4+1"-" 2)=5/3.

答案5/3

6.导学号93774090求证:sinx/(1+cosx)-cosx/(1+sinx)=(2"(" sinx"-" cosx")" )/(1+sinx+cosx).

证明方法一:左边=(sinx+sin^2 x"-" cosx"-" cos^2 x)/("(" 1+cosx")(" 1+sinx")" )

　　=("(" sinx"-" cosx")(" 1+sinx+cosx")" )/(1+sinx+cosx+cosx"·" sinx)

　　=(2"(" sinx"-" cosx")(" 1+sinx+cosx")" )/(1+sin^2 x+cos^2 x+2sinx+2cosx+2cosx"·" sinx)

　　=(2"(" sinx"-" cosx")(" 1+sinx+cosx")" )/("(" 1+sinx+cosx")" ^2 )

　　=(2"(" sinx"-" cosx")" )/(1+sinx+cosx)

　　=右边.

　　方法二:左边

　　=(1+sinx+cosx)/(1+sinx+cosx) (sinx/(1+cosx) "-" cosx/(1+sinx))

　　=1/(1+sinx+cosx) [(sinx"(" 1+sinx+cosx")" )/(1+cosx)┤-├ (cosx"(" 1+sinx+cosx")" )/(1+sinx)]

　　=1/(1+sinx+cosx) (sinx+(sin^2 x)/(1+cosx)┤-├ cosx"-" (cos^2 x)/(1+sinx))

　　=1/(1+sinx+cosx)(sin x+1-cos x-cos x-1+sin x)

　　=(2"(" sinx"-" cosx")" )/(1+sinx+cosx)

　　=右边.

7.导学号93774091已知sin α,cos α是方程5x2-x+m=0的两个实根.

(1)求m的值;

(2)当α∈(0,π)时,求tan(3π-α)的值;

(3)求sin3α+cos3α的值.

解(1)∵sin α,cos α是5x2-x+m=0的两个实根,

　　∴{■(sinα+cosα=1/5 "," @sinαcosα=m/5 "." )┤

又sin2α+cos2α=1,