答案 B
解析 a =2sin(45°+14°)=2sin59°
b =2sin(45°+16°)=2sin61°
c =
6
2
=2sin60°,∴b >c >a . 6.在△ABC 中,C =120°,tan A +tan B =23
3,则cos A cos B =( )
A.14
B.34
C.12 D .-14
答案 B
解析 tan A +tan B =sin A cos A +sin B
cos B
=sin A cos B +cos A sin B cos A cos B =sin A +B cos A cos B =sin60°cos A cos B =32cos A cos B =23
3
∴cos A cos B =3
4
7.已知tan(α+β)=25,tan ? ????β-π4=14,那么tan ? ????α+π4等于( ) A.13
18 B.1322 C.322
D.16
答案 C
解析 因为α+π4+β-π4=α+β,所以α+π4=(α+β)-?
????β-π4,所以
tan ? ????α+π4=tan ?
???
??
α+β-?
????β-π4
=tan α+β-tan ?
????β-π41+tan α+βtan ? ????β-π4=322.
8.(09·陕西卷)若3sin α+cos α=0,则1
cos 2
α+sin2α
的值为( )
A.10
3
B.53
C.23 D .-2
答案 A
解析 3sin α=-cos α?tan α=-1
3
.
1cos 2α+sin2α=cos 2α+sin 2αcos 2
α+2sin αcos α=1+tan 2
α
1+2tan α=1+191-23=103
. 二、填空题
9.cos84°cos24°-cos114°cos6°的值为________. 答案 12
解析 cos84°cos24°-cos114°cos6°=cos84°cos24°+cos66°sin84°=cos84°cos24°+sin24°sin84°=cos(84°-24°)=cos60°=1
2
.
10.(2010·全国卷Ⅰ,理)已知α为第三象限的角,cos 2α=-35,则tan (π
4+2α)
=________.
答案 -1
7
解析 由cos 2α=2cos 2
α-1=-35,且α为第三象限角,得cos α=-55,sin α=
-
255,则tan α=2,tan2α=-43,tan(π4+2α)=1+tan 2α1-tan 2α=-1
7. 11.(2011·潍坊)化简:
sin 3α-πsin α+cos 3α-π
cos α
=________.
答案 -4cos2α
解析 原式=-sin3αsin α+-cos3α
cos α=
-sin3αcos α+cos3αsin αsin αcos α=-sin4α
sin αcos α=
-
4sin αcos α·cos2α
sin αcos α
=-4cos2α.
12.不查表,计算1sin10°-3
sin80°=________.(用数字作答)
答案 4
解析 原式=cos10°-3sin10°
sin10°cos10°
=212cos10°-3
2
sin10°sin10°cos10°
=4
sin30°cos10°-cos30°sin10°
2sin10°cos10°
=4sin 30°-10°
sin20°
=4.
三、解答题
13.求(tan10°-3)·cos10°
sin50°的值.
解析 (tan10°-3)·cos10°sin50°=(tan10°-tan60°)·cos10°sin50°=(sin10°
cos10°-
sin60°cos60°
)·
cos10°sin50°
=
sin10°cos60°-sin60°cos10°
cos10°cos60°
·
cos10°sin50°
=
-sin 60°-10°cos10°·cos60°·cos10°sin50°=-1
cos60°
=-2.
14.已知sin(α+π4)=45,且π4<α<3π
4.求cos α的值.
解析 sin(α+π4)=45且π4<α<3π
4
∴
π
2<α+π
4
<π
∴cos(α+π
4
)=-
1-sin
2
α+π4
=-35
∴cos α=cos[(α+π4)-π
4
]
=cos(α+π4)cos π4+sin(α+π4)sin π
4
=-35×22+45×22=2
10
.
15.已知tan2θ=34(π
2<θ<π),求2cos 2
θ
2+sin θ-1
2cos θ+
π
4的值.
解 ∵tan2θ=
2tan θ1-tan 2
θ=3
4
, ∴tan θ=-3或tan θ=1
3,
又θ∈(π
2,π),∴tan θ=-3,
∴
2cos 2
θ
2
+sin θ-1
2cos θ+
π4=cos θ+sin θcos θ-sin θ=1+tan θ1-tan θ
=1-31+3=-1
2
.