人教B 版必修4同步练习
1.若M =cos17°sin13°+sin17°cos13°,则M 的值为( ) A.12 B.22 C.32
D. 以上都不对 解析:选A.原式=sin(13°+17°)=sin30°=1
2
.
2.sin65°cos35°-cos65°sin35°等于( ) A.12 B.32
C .-32
D .-1
2
解析:选A.原式=sin(65°-35°)=sin30°=1
2
.
3.若M =sin12°cos57°-cos12°sin57°,N =cos10°cos55°+sin10°sin55°,则以下判断正确的是( )
A .M >N
B .M =N
C .M +N =0
D .MN =1
2
解析:选C.M =sin(12°-57°)=sin(-45°)=-sin45°=-2
2
,
N =cos(10°-55°)=cos(-45°)=cos45°=2
2
,
∴M +N =0.
4.化简:sin(α+β)+sin(α-β)+2sin αsin ????
3π2-β=________. 解析:原式=2sin αcos β-2sin αcos β=0. 答案:0
一、选择题
1.(2010年高考福建卷)计算sin43°·cos13°-cos43°·sin13°的结果等于( ) A.12 B.33 C.22 D.32
解析:选A.原式=sin(43°-13°)=sin30°=1
2
.
2.(2011年金华高一检测)已知A (3,0),B (0,3),C (cos α,sin α),若AC →·BC →
=-1,则sin ????α+π4等于( )
A.13
B.
23 C.33
D.23 解析:选B.AC →=(cos α-3,sin α),BC →
=(cos α,sin α-3), ∴AC →·BC →=(cos α-3)cos α+sin α(sin α-3)
=cos 2α-3cos α+sin 2α-3sin α =1-3(sin α+cos α)=-1,
∴3(sin α+cos α)=2,
∴32sin(α+π
4)=2,
∴sin(α+π4)=2
3
.
3.在△ABC 中,若sin A cos B =1-cos A sin B ,则△ABC 一定是( ) A .锐角三角形 B .直角三角形 C .钝角三角形 D .等腰三角形 解析:选B.∵sin A cos B =1-cos A sin B , ∴sin A cos B +cos A sin B =1, 即sin(A +B )=1. ∵A ,B 为三角形的内角, ∴A +B =90°, ∴∠C =90°,
∴△ABC 为直角三角形.
4.在△ABC 中,A =π4,cos B =1010
,则sin C =( )
A .-55 B.5
5
C .-255 D.255
解析:选D.∵cos B =1010∴sin B =310
10,
∴sin C =sin(A +B )=sin A cos B +cos A sin B
=
22×1010+22×31010=25
5
. 5.若0<α<β<π
4
,sin α+cos α=a ,sin β+cos β=b ,则( )
A .a >b
B .a <b
C .ab <1
D .ab >2
解析:选B.a =2sin(α+π
4
),b =2sin ????β+π4. f (x )=2sin ????x +π4在????0,π
4上是增函数. 又0<α<β<π
4,
∴f (α)<f (β),即a <b .
6.设△ABC 的三个内角为A ,B ,C ,向量m =(3sin A ,sin B ),n =(cos B ,3cos A ),若m·n =1+cos(A +B ),则C 等于( )
A.π6
B.π3
C.2π3
D.5π6
解析:选C.∵m ·n =1+cos(A +B ) =3sin A cos B +3cos A sin B ,
∴3sin(A +B )=1+cos(A +B ).
又A +B =π-C ,∴整理得sin(C +π6)=1
2
,
∵0<C <π,∴π6<C +π6<7π
6
,
∴C +π6=5π6,∴C =2π3.
二、填空题
7.函数f (x )=3sin x +sin ????
π2+x 的最大值是________.
解析:f (x )=3sin x +cos x =2sin ????x +π6, ∴f (x )的最大值为2.
答案:2
8.sin(x +60°)+2sin(x -60°)-3cos(120°-x )=________.
解析:原式=sin x cos60°+cos x sin60°+2sin x cos60°-2cos x sin60°-3(cos120°cos x +sin120°sin x )
=32sin x -32cos x +32cos x -3
2sin x =0. 答案:0 9.cos10°tan20°
+3sin10°tan70°-2cos40°=________. 解析:cos10°tan20°+3sin10°tan70°-2cos40°
=cos20°cos10°sin20°+3sin10°sin70°cos70°
-2cos40°
=
cos20°cos10°+3sin10°cos20°
sin20°
-2cos40°
=cos20°(cos10°+3sin10°)sin20°
-2cos40°
=2cos20°(cos10°sin30°+sin10°cos30°)sin20°-2cos40°
=2cos20°sin40°-2sin20°cos40°sin20°
=2. 答案:2 三、解答题
10.已知π4<α<3π4,0<β<π4,cos ????π4+α=-35
, sin ????3π4+β=513,求sin(α+β)的值.
解:∵π4<α<34π,π2<π
4+α<π,
∴sin ????π4+α=1-cos 2????π4+α=45. ∵0<β<π4,34π<3
4π+β<π,
∴cos ???
?3
4π+β
=-
1-sin 2????34π+β=-1213
, ∴sin(α+β)=-sin(π+α+β)
=-sin ???
?????π4+α+????34π+β =-???
?sin ????π4+αcos ????34π+β+cos ????π4+αsin ????3π4+β =-????45×????-1213+????-35×513=6365
. 11.设A ,B 为锐角三角形ABC 的两个内角,向量a =(2cos A,2sin A ),b =(3cos B,3sin B ),若a ,b 的夹角为60°,求A -B 的值.
解:∵|a |=2,|b |=3,
a ·
b =2cos A ·3cos B +2sin A ·3sin B =6(cos A cos B +sin A sin B )=6cos(A -B ) 而a 与b 的夹角为60°,
则cos60°=12=a ·b
|a |·|b |
=6cos (A -B )
2×3
=cos(A -B )
即cos(A -B )=1
2.
又∵0<A <π2,0<B <π
2
,
∴-π2<A -B <π2
,
∴A -B =±π
3
.
12.设函数f (x )=a·b ,其中向量a =(m ,cos2x ),b =(1+sin2x,1),x ∈R ,且y =f (x )的
图象经过点(π
4
,2).
(1)求实数m 的值;
(2)求函数f (x )的最小值及此时x 值的集合.
解:(1)f (x )=a·b =m (1+sin2x )+cos2x
因为f (x )图象经过点(π
4
,2).
∴f (π4)=2,即m (1+sin π2)+cos π
2=2, ∴m =1.
(2)由(1)得f (x )=1+sin2x +cos2x
=1+2sin(2x +π
4).
故当sin(2x +π
4)=-1时,f (x )取得最小值,
f (x )min =1- 2.
相应的2x +π4=3
2π+2k π,k ∈Z ,
∴x =k π+5
8
π,k ∈Z ,
∴使函数f (x )取得最小值的x 的集合为{x |x =k π+5
8
π,k ∈Z }.