1.已知cos2θ=
2
3
,则sin 4θ+cos 4θ的值为( ) A.1318 B.1118 C.79
D .-1 解析:选B.sin 4θ+cos 4θ=(sin 2θ+cos 2θ)2-2sin 2θcos 2θ
=1-12sin 22θ=1-12(1-cos 22θ)=1118
.
2.(2012·绵阳调研)已知α是锐角,且sin(π2+α)=34,则sin(α
2
+π)的值等于( )
A.24 B .-24 C.144 D .-144
解析:选B.由sin(π2+α)=34,得cos α=3
4
,
又α为锐角,
∴sin(α2+π)=-sin α
2=-1-cos α2
=-1-342=-18=-2
4
.
3.化简sin 235°-
1
2
cos10°cos80°
=( )
A .-2
B .-1
2
C .-1
D .1
解析:选C.sin 235°-12cos10°cos80°=1-cos70°2-12cos10°sin10°=-1
2
cos70°
1
2
sin20°=-1.故选C.
4.已知cos(α+β)+cos(α-β)=45,sin(α+β)+sin(α-β)=3
5
.求:
(1)tan α;(2)2cos 2α
2
-3sin α-1
2sin ???
?α+π4.
解:(1)由已知得2cos αcos β=4
5
.①
2sin αcos β=3
5
.②
②÷①得,tan α=3
4
.
(2)原式=cos α-3sin αsin α+cos α=1-3tan α
1+tan α,
由(1)得tan α=3
4
,代入上式得
2cos 2α2-3sin α-12sin ???
?α+π4=1-3×
341+34=-5
7.
一、选择题
1.在△ABC 中,若cos2B +3cos(A +C )+2=0,则sin B 的值是( ) A.12 B.22 C.32
D .1 解析:选C.由cos2B +3cos(A +C )+2=0, 得2cos 2B -3cos B +1=0,
所以cos B =1
2,或cos B =1(舍去),
∴sin B =3
2
.
2.已知tan α=-13,且-π
2<α<0,则2sin 2α+sin2αcos ???
?α-π4=( )
A .-255
B .-3510
C .-31010 D.255
解析:选A.因tan α=-1
3
,
又-π2<α<0,所以sin α=-1010.
故2sin 2α+sin2αcos ????α-π4=2sin α(sin α+cos α)22(sin α+cos α)
=22sin α=-25
5
.
3.(2012·宜昌调研)已知角A 为△ABC 的内角,且sin2A =-3
4
,则sin A -cos A =( )
A.72 B .-72 C .-12 D.12
解析:选A.∵A 为△ABC 的内角且sin2A =2sin A cos A =-3
4
<0,
∴sin A >0,cos A <0, ∴sin A -cos A >0.
又(sin A -cos A )2=1-2sin A cos A =7
4
,
∴sin A -cos A =7
2
.
4.(2010·高考课标全国卷)若cos α=-4
5,α是第三象限的角,则1+tan
α21-tan
α
2
=( )
A .-12 B.12
C .2
D . -2
解析:选A.∵α是第三象限角,cos α=-4
5
,
∴sin α=-3
5
.
∴1+tan α21-tan α2=1+
sin α2cos α21-
sin α2cos α2
=cos α2+sin α2
cos α2-sin
α2
=cos α2+sin α2cos α2-sin α2·cos α2+sin α2cos α2+sin α2
=1+sin αcos α=1-35-45=-1
2
.
5.tan70°·cos10°(3tan20°-1)等于( ) A .1 B .2 C .-1 D .-2 解析:选C.tan70°·cos10°(3tan20°-1) =sin70°cos70°·cos10°(3·sin20°cos20°
-1) =cos20°cos10°sin20°·3sin20°
-cos20°cos20°
=cos10°·2sin (20°-30°)sin20°=-sin20°sin20°
=-1.
二、填空题
6.若cos(α+β)=15,cos(α-β)=3
5
,则tan α·tan β=________.
解析:∵cos(α+β)=cos αcos β-sin αsin β=1
5
,①
cos(α-β)=cos αcos β+sin αsin β=3
5
.②
由①②解得cos αcos β=25,sin αsin β=1
5
,
则tan αtan β=sin αsin βcos αcos β=1
2.
答案:12
7.已知sin 2(2x -π4)=1
4
,则sin4x =________.
解析:sin 2????2x -π4=1-cos ????4x -π22=12-12sin4x =1
4, ∴sin4x =1
2.
答案:12
8.若α=20°,β=25°,则(1+tan α)(1+tan β)的值为________.
解析:由tan(α+β)=tan α+tan β
1-tan αtan β
=tan45°=1可得tan α+tan β+tan αtan β=1,
所以(1+tan α)(1+tan β)=1+tan α+tan β+tan αtan β=2. 答案:2 三、解答题
9.(2012·荆州质检)已知向量a =(sin θ,2),b =(cos θ,1),且a ∥b ,其中θ∈????0,π2. (1)求sin θ和cos θ的值;
(2)若sin(θ-ω)=35,0<ω<π
2
,求cos ω的值.
解:(1)∵a =(sin θ,2),b =(cos θ,1),且a ∥b , ∴sin θ2=cos θ1
,即sin θ=2cos θ.
又∵sin 2θ+cos 2θ=1,θ∈???
?0,π2, ∴sin θ=255,cos θ=5
5.
(2)∵0<ω<π2,0<θ<π2,∴-π2<θ-ω<π
2.
∵sin(θ-ω)=3
5
,
∴cos(θ-ω)=1-sin 2(θ-ω)=4
5
.
∴cos ω=cos[θ-(θ-ω)]=cos θcos(θ-ω)+sin θsin(θ-ω)=25
5
.
10.已知cos α=17,cos(α-β)=1314,且0<β<α<π
2
.
(1)求tan2α的值; (2)求β.
解:(1)由cos α=17,0<α<π
2
,
得sin α=1-cos 2α= 1-(17)2=43
7
.
∴tan α=sin αcos α=437×7
1
=4 3.
于是tan2α=2tan α1-tan 2α=2×431-(43)2
=-83
47. (2)由0<β<α<π2,得0<α-β<π
2.
又∵cos(α-β)=13
14
,
∴sin(α-β)=1-cos 2(α-β)= 1-(1314)2=33
14
.
由β=α-(α-β),
得cos β=cos[α-(α-β)]=cos αcos(α-β)+sin αsin(α-β)
=17×1314+437×3314=12. ∴β=π3
.
11.已知tan(π+α)=-1
3,tan(α+β)=sin2????π
2-α+4cos 2α10cos 2α-sin2α
.
(1)求tan(α+β)的值; (2)求tan β的值.
解:(1)∵tan(π+α)=-13,∴tan α=-1
3.
∵tan(α+β)=sin2???
?π
2-α+4cos 2α10cos 2α-sin2α
=sin2α+4cos 2α10cos 2α-sin2α=2sin αcos α+4cos 2α10cos 2α-2sin αcos α =2cos α(sin α+2cos α)2cos α(5cos α-sin α) =sin α+2cos α5cos α-sin α=tan α+25-tan α =-13+25-???
?-13=516.
(2)tan β=tan[(α+β)-β]=tan (α+β)-tan α
1+tan (α+β)tan α
=516+131-516×13
=3143.