课时跟踪检测(二十四) 两角差的余弦函数两角和与差的正弦、余弦函数一、基本能力达标1.已知α∈⎝ ⎛⎭⎪⎫0,π2,cos α=33,则cos ⎝⎛⎭⎪⎫α+π6=( )A.12-66 B .1-66 C .-12+66 D .-1+66解析:选A ∵α∈⎝ ⎛⎭⎪⎫0,π2,cos α=33,∴sin α=63,∴cos ⎝⎛⎭⎪⎫α+π6=cos αcos π6-sin αsin π6=33×32-63×12=12-66. 2.满足cos αcos β=32-sin αsin β的一组α,β的值是 ( )A .α=13π12,β=3π4B .α=π2,β=π3C .α=π2,β=π6D .α=π3,β=π4解析:选B ∵cos αcos β=32-sin αsin β,∴cos αcos β+sin αsin β=32,即cos(α-β)=32,经验证可知选项B 正确.3.在△ABC 中,若sin A sin B <cos A cos B ,则△ABC 一定是 ( ) A .直角三角形 B .锐角三角形 C .钝角三角形 D .三者都有可能 解析:选C ∵sin A sin B <cos A cos B , ∴cos A cos B -sin A sin B >0,∴cos(A +B )>0,∴A +B <90°,∴C >90°,∴△ABC 是钝角三角形.4.已知3cos x -sin x =-65,则sin ⎝ ⎛⎭⎪⎫π3-x = ( )A.45 B .-45 C.35 D .-35 解析:选D3cos x -sin x =2⎝ ⎛⎭⎪⎫sin π3cos x -cos π3sin x=2sin ⎝ ⎛⎭⎪⎫π3-x =-65,故sin ⎝ ⎛⎭⎪⎫π3-x =-35.5.已知0<α<π2<β<π,又sin α=35,sin(α+β)=35,则sin β等于( )A .0B .0或2425 C.2425 D .±2425解析:选C 由0<α<π2<β<π得,π2<α+β<3π2,又sin α=35,sin(α+β)=35,∴cos α=45,cos(α+β)=-45,∴sin β=sin[(α+β)-α]=sin(α+β)cos α-cos(α+β)sin α=35×45-⎝ ⎛⎭⎪⎫-45×35=2425.6.sin 15°+cos 165°的值是________. 解析:原式=sin(45°-30°)+cos(120°+45°)=sin 45°cos 30°-cos 45°sin 30°+cos 120°cos 45°-sin 120°sin 45°=22×32-22×12-12×22-32×22=-22.答案:-227.设a =2cos 66°,b =cos 5°-3sin 5°,c =2(sin 47°sin 66°-sin 24°sin 43°),则a ,b ,c 的大小关系是________. 解析:∵b =2cos 65°,c =2(cos 43°cos 24°-sin 24°sin 43°)=2cos 67°,∴b >a >c .答案:b >a >c8.已知sin α+sin β+sin γ=0和cos α+cos β+cos γ=0,则cos(α-β)的值是________. 解析:由已知得,-sin γ=sin α+sin β, ① -cos γ=cos α+cos β, ②①2+②2得,1=1+1+2sin αsin β+2cos αcos β,化简得cos αcos β+sin αsin β=-12,即cos(α-β)=-12.答案:-129.已知cos(α-β)=-1213,cos(α+β)=1213,且α-β∈⎝ ⎛⎭⎪⎫π2,π,α+β∈⎝ ⎛⎭⎪⎫3π2,2π,求角β的值.解:由α-β∈⎝ ⎛⎭⎪⎫π2,π,且cos(α-β)=-1213,得sin(α-β)=513. 由α+β∈⎝ ⎛⎭⎪⎫3π2,2π,且cos(α+β)=1213,得sin(α+β)=-513,cos 2β=cos[(α+β)-(α-β)]=cos(α+β)cos(α-β)+sin(α+β)sin(α-β) =1213×⎝ ⎛⎭⎪⎫-1213+⎝ ⎛⎭⎪⎫-513×513=-1.又∵α-β∈⎝ ⎛⎭⎪⎫π2,π,α+β∈⎝⎛⎭⎪⎫3π2,2π, ∴2β∈⎝ ⎛⎭⎪⎫π2,3π2,∴2β=π,则β=π2.10.已知π4<α<3π4,0<β<π4,cos ⎝ ⎛⎭⎪⎫π4+α=-35,sin ⎝⎛⎭⎪⎫3π4+β=513,求sin(α+β). 解:∵π4<α<3π4,0<β<π4,∴π2<π4+α<π,3π4<3π4+β<π,又cos ⎝ ⎛⎭⎪⎫π4+α=-35,sin ⎝⎛⎭⎪⎫3π4+β=513, ∴sin ⎝ ⎛⎭⎪⎫π4+α=45,cos ⎝⎛⎭⎪⎫3π4+β=-1213, ∴sin(α+β)=-sin ⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫π4+α+⎝⎛⎭⎪⎫3π4+β =-⎣⎢⎡⎦⎥⎤sin ⎝ ⎛⎭⎪⎫π4+αcos ⎝⎛⎭⎪⎫3π4+β+cos ⎝ ⎛⎭⎪⎫π4+αsin ⎝ ⎛⎭⎪⎫3π4+β =-⎣⎢⎡⎦⎥⎤45×⎝ ⎛⎭⎪⎫-1213+⎝ ⎛⎭⎪⎫-35×513=6365.二、综合能力提升1.在△ABC 中,A =π4,cos B =1010,则sin C = ( )A .-55 B.55 C .-255 D.255解析:选D ∵A =π4,∴cos A =sin A =22,又cos B =1010,0<B <π,∴sin B =31010,∴sin C =sin(A +B )=sin A cos B +cos A sin B =22×1010+22×31010=255. 2.已知sin(α+β)=14,sin(α-β)=13,则tan αtan β的值为( )A .-17 B.17 C .-7 D .7解析:选C 由sin(α+β)=14得sin αcos β+cos αsin β=14,①由sin(α-β)=13得sin αcos β-cos αsin β=13,②由①②得sin αcos β=724,cos αsin β=-124,以上两式相除得tan αtan β=-7.3.在△ABC 中,若2cos B sin A =sin C ,则△ABC 的形状一定是 ( ) A .等腰直角三角形 B .直角三角形C .等腰三角形 D .等边三角形 解析:选C ∵在△ABC 中,A +B +C =π,∴C =π-(A +B ).∴sin C =sin(A +B )=sin A cos B +cos A sin B , ∴2cos B sin A =sin A cos B +cos A sin B ,即sin A cos B -cos A sin B =0.∴sin(A -B )=0,∴A =B .4.若cos αcos β=32-sin αsin β,且α∈⎝ ⎛⎭⎪⎫0,π2,β∈⎝ ⎛⎭⎪⎫π2,π,则α-β的值是 ( )A .-π6B .-π3 C.π6 D.π3解析:选A 由题意知cos(α-β)=32,又0<α<π2,-π<-β<-π2,所以-π<α-β<0,故α-β=-π6.5.已知cos ⎝ ⎛⎭⎪⎫x -π6=-33,则cos x +cos ⎝ ⎛⎭⎪⎫x -π3=________.解析:cos x +cos ⎝⎛⎭⎪⎫x -π3=cos x +12cos x +32sin x=32cos x +32sin x =3⎝ ⎛⎭⎪⎪⎫32cos x +12sin x =3cos ⎝ ⎛⎭⎪⎫x -π6=-1.答案:-16.已知cos(α+β)=45,cos(α-β)=-45,则cos αcos β=________.解析:由条件知:⎩⎪⎨⎪⎧cos (α+β)=cos αcos β-sin αsin β=45, ①cos (α-β)=cos αcos β+sin αsin β=-45. ②①+②得2cos αcos β=0,∴ cos αcos β=0. 答案:07.已知向量a =(cos α,sin α),b =(cos β,sin β),|a -b |=255,求cos(α-β)的值.解:∵a =(cos α,sin α),b =(cos β,sin β), ∴a -b =(cos α-cos β,sin α-sin β). ∴|a -b |=(cos α-cos β)2+(sin α-sin β)2=cos 2α-2cos αcos β+cos 2β+sin 2α-2sin αsin β+sin 2β =2-2cos (α-β)=255,∴2-2cos(α-β)=45,∴cos(α-β)=35.8.已知cos α=55,sin(α-β)=1010,且α,β∈⎝⎛⎭⎪⎫0,π2.求:(1)cos(2α-β)的值; (2)β的值.解:(1)因为α,β∈⎝ ⎛⎭⎪⎫0,π2,所以α-β∈⎝ ⎛⎭⎪⎫-π2,π2,又sin(α-β)=1010>0,∴0<α-β<π2.所以sin α= 1-cos 2α=255,cos(α-β)= 1-sin 2(α-β)=31010,cos(2α-β)=cos[α+(α-β)] =cos αcos(α-β)-sin αsin(α-β)=55×31010-255×1010=210. (2)cos β=cos[α-(α-β)]=cos αcos(α-β)+sin αsin(α-β) =55×31010+255×1010=22, 又因为β∈⎝⎛⎭⎪⎫0,π2,所以β=π4.。