第3模块 第2节
[知能演练]
一、选择题
1.α是第四象限角,tan α=-512
,则sin α等于 ( )
A.15
B .-15C.513
D .-513解析:?????
sin αcos α=-512,sin 2α+cos 2α=1,
∴??? sin α=
513,cos α=-1213或??? sin α=-513,cos α=1213. ∵α是第四象限角,∴sin α<0,cos α>0. ∴sin α=-
513
.选D. 答案:D 2.已知cos(π2+φ)=32,且|φ|<π2
,则tan φ等于 ( )
A .-33
B.33 C .- 3 D. 3
解析:由cos(π2φ)=32,得sin φ=-32
. 又|φ|<π2,∴cos φ=12
.∴tan φ=- 3. 答案:C
3.若α是第三象限角,且cos(75°+α)=13
,则tan(15°-α)的值为 ( )
A .-
223 B .-24 C.223
D.24
解析:cos(75°+α)=sin(90°-75°-α)=sin(15°-α)=13
,又∵α为第三象限角, ∴-α为第二象限角.
∴-α+15°为第二象限角.
∴cos(15°-α)=-
1-19=-223. ∴tan(15°-α)=-
24. 答案:B
4.若△ABC 的内角A 满足sin2A =23
,则sin A +cos A 等于 ( )
A.153
B .-153 C.53 D .-53
解析:在△ABC 中,2sin A cos A =23
>0, ∴sin A >0,cos A >0. ∴sin A +cos A =(sin A +cos A )2
=sin 2A +cos 2A +2sin A cos A =1+23
=53=153. 答案:A
二、填空题 5.如果cos α=15,且α是第四象限角,那么cos(α+π2
)=________. 解析:由已知?cos(α+π2)=-sin α=-(-1-cos 2α)=265
. 答案:265
6.化简:sin 2(α+π)·cos(π+α)·cos(-α-2π)tan(π+α)·sin 3(π2
+α)·sin(-α-2π)=
________.
解析:sin 2
(α+π)·cos(π+α)·cos(-α-2π)tan(π+α)·sin 3(π2
+α)·sin(-α-2π)
=(-sin α)2·(-cos α)·cos(-α)tan α·cos 3α·sin(-α)
=-sin 2α·cos α·cos αsin αcos α
·cos 3α·(-sin α)=sin 2αcos 2αsin 2αcos 2α=1. 答案:1
三、解答题
7.已知cos(π+α)=-12
,且α是第四象限角,计算: (1)sin(2π-α);
(2)sin[α+(2n +1)π]+sin[α-(2n +1)π]sin(α+2nπ)·cos(α-2nπ)
(n ∈Z). 解:∵cos(π+α)=-12,∴-cos α=-12,cos α=12
, 又∵α是第四象限角,
∴sin α=-1-cos 2α=-
32
. (1)sin(2π-α)=sin[2π+(-α)] =sin(-α)=-sin α=
32
. (2)sin[α+(2n +1)π]+sin[α-(2n +1)π]sin(α+2nπ)·cos(α-2nπ) =sin(2nπ+π+α)+sin(-2nπ-π+α)sin(2nπ+α)·cos(-2nπ+α)
=sin(π+α)+sin(-π+α)sin α·cos α
=-sin α-sin(π-α)sin α·cos α=-2sin αsin αcos α
=-2cos α
=-4. 8.已知sin(π-α)-cos(π+α)=
23(π2<α<π).求下列各式的值: (1)sin α-cos α;
(2)sin 3(π2-α)+cos 3(π2
+α). 解:由sin(π-α)-cos(π+α)=
23
, 得sin α+cos α=23.① 将①式两边平方,得1+2sin α·cos α=29
, 故2sin α·cos α=-79
, 又π2
<α<π,∴sin α>0,cos α<0. ∴sin α-cos α>0.
(1)(sin α-cos α)2=1-2sin α·cos α
=1-(-79)=169
,
∴sin α-cos α=43
. (2)sin 3(π2-α)+cos 3(π2
+α)=cos 3α-sin 3α =(cos α-sin α)(cos 2α+cos α·sin α+sin 2α)
=(-43)×(1-718=-2227
[高考·模拟·预测]
1.已知tan θ=2,则sin 2θ+sin θcos θ-2cos 2θ=
( )
A .-43
B.54 C .-34 D.45
解析:由于tan θ=2,则sin 2θ+sin θcos θ-2cos 2θ
=sin 2θ+sin θcos θ-2cos 2θsin 2θ+cos 2θ=tan 2θ+tan θ-2tan 2θ+1
=22
+2-222+1=45
,故选D. 答案:D
2.已知△ABC 中,1tan A =-125
,则cos A = ( )
A.1213
B.513 C .-
513 D .-1213