2018_2019学年高中数学第三章三角恒等变换3.1两角和与差的正弦余弦和正切公式2课后习题新人教A版必修4.doc

3.1.2两角和与差的正弦、余弦、正切公式课后篇巩固探究A组基础巩固1.若sin=cos,则tan =()A.-1B.0C.D.1解析由已知得cos -sin =cos -sin ,因此sin =cos ,于是tan =-1.答案A2.已知a=(2sin 35,2cos 35),b=(cos 5,-sin 5),则ab=()A.B.1C.2D.2sin 40解析ab=2sin 35cos 5-2cos 35sin 5=2sin(35-5)=2sin 30=1.答案B3.若tan(+)=,tan(-)=,则tan 2=()A.B.C.D.解析tan 2=tan (+)+(-)=.答案D4.sin(+75)+cos(+45)-cos(+15)的值等于()A.1B.1C.-1D.0解析原式=sin (+45)+30+cos(+45)-cos (+45)-30=sin(+45)+cos(+45)+cos(+45)-=sin(+45)+cos(+45)-cos(+45)-sin(+45)=0.答案D5.设,且tan =,则()A.3-=B.3+=C.2-=D.2+=解析由tan =,得,得sin cos -cos sin =cos ,sin(-)=sin.又,故-=-,即2-=.答案C6.化简:=.解析原式=-1.答案-17.已知tan(+)=,tan,则tan的值为.解析因为tan(+)=,tan,所以tan=tan =.答案8.若是锐角,且满足sin,则cos 的值为.解析是锐角,-.又sin,cos.cos =cos=coscos-sinsin=.答案9.tan 23+tan 37+tan 23tan 37的值是.解析tan 60=,tan 23+tan 37=tan 23tan 37,tan 23+tan 37+tan 23tan 37=.答案10.化简求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-sin(70+)cos(10+);(3)cos 21cos 24+sin 159sin 204.解(1)原式=sin(+-)=sin 2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin (10+)-(70+)=sin(-60)=-.(3)原式=cos 21cos 24+sin(180-21)sin(180+24)=cos 21cos 24-sin 21sin 24=cos(21+24)=cos 45=.11.已知cos =-,tan =,0,求-的值.解法一由cos =-,得sin =-,tan =2,又tan =,于是tan(-)=1.又由,0,可得-0,-,因此-=.解法二由cos =-,得sin =-.由tan =,0,得sin =,cos =.所以sin(-)=sin cos -cos sin =-.又由,0,可得-0,-,因此,-=.B组能力提升1.已知,tan=-3,则sin =()A.B.-C.D.解析tan =tan =-,因为,所以,故sin =.答案A2.导学号68254102设,都为锐角,且cos =,sin(+)=,则sin 等于()A.B.C.D.-解析为锐角,cos =,sin =.,都为锐角,0+.sin(+)=,cos(+)=.当cos(+)=-时,sin =sin(+)-=sin(+)cos -cos(+)sin =;当cos(+)=时,sin =sin(+)-=sin(+)cos -cos(+)sin =-,与已知为锐角矛盾.sin =.答案B3.若将函数f(x)=sin 2x+cos 2x的图象向右平移个单位长度,所得图象关于y轴对称,则的最小正值是()A.B.C.D.解析由题意f(x)=sin 2x+cos 2x=sin,将其图象向右平移个单位长度,得函数y=sinsin的图象,要使图象关于y轴对称,则-2=+k,解得=-,当k=-1时,取最小正值.答案C4.已知cos(+)=,cos(-)=-,则cos cos =.解析由已知得cos cos -sin sin =,cos cos +sin sin =-,两式相加得2cos cos =0,故cos cos =0.答案05.已知ABC中,tan Atan B-tan A-tan B=,则C的大小为.解析依题意,=-,即tan(A+B)=-,又0A+B,所以A+B=,故C=-A-B=.答案6.已知,均为锐角,且tan =,求tan(+)的值.解tan =tan,因为,均为锐角,所以-,0,又y=tan x在上是单调函数,所以=-,即+=,tan(+)=1.7.导学号68254103已知向量a=(cos ,sin ),b=(cos ,sin ),|a-b|=.(1)求cos(-)的值;(2)若-0,且sin =-,求sin 的值.解(1)a=(cos ,sin ),b=(cos ,sin ),|a|=|b|=1,|a-b|2=a2-2ab+b2=1+1-2(cos cos +sin sin )=2-2cos(-).又|a-b|=,|a-b|2=2-2cos(-)=,cos(-)=.(2)-0,0-0)有最大值1和最小值-4,求a,b的值.解f(x)=(cos x-sin x)sin-2asin x+b=(cos2x-sin2x)-2asin x+b=(1-2sin2x)-2asin x+b=-(sin x+a)2+a2+b.当a1时,f(x)的最小值等于f,最大值等于f,依题意得解得a=,b=-1.当0a1时,依题意可得解得a=-1(舍去)或a=-1(舍去).综上可得a=,b=-1.7