2018-2019全国高中数学第三章三角恒等变换3.3三角函数的积化和差与和差化积练习新人教B版.docx

3.3三角函数的积化和差与和差化积课时过关能力提升1.式子sin 15sin 105的值等于()A.B.-C.D.-解析:sin 15sin 105=- cos 120-cos(-90)=-.答案:A2.式子sin 20+cos 10可化简为()A.sin 50B.cos 50C.sin 50D.cos 50解析:sin 20+cos 10=sin 20+sin 80=2sin 50cos 30=sin 50.答案:C3.若cos(+)cos(-)=,则cos2-sin2等于()A.-B.-C.D.解析:由cos(+)cos(-)=(cos cos -sin sin )(cos cos +sin sin )=cos2cos2-sin2sin2=cos2(1-sin2)-sin2sin2=cos2-cos2sin2-sin2sin2=cos2-sin2(cos2+sin2)=cos2-sin2,知cos2-sin2=.答案:C4.已知直角三角形中两锐角为A和B,则sin Asin B()A.有最大值和最小值0B.有最大值,但无最小值C.既无最大值,也无最小值D.有最大值1,但无最小值解析:sin Asin B=- cos(A+B)-cos(A-B)=-cos(A-B).又0A,0B,所以-A-B,于是0cos(A-B)1,故0sin Asin B.答案:B5.cos 72-cos 36等于()A.3-2B.C.-D.-解析:cos 72-cos 36=-2sin 54sin 18=-2sin 18cos 36=-=-.答案:C6.已知-=,且cos -cos =,则cos(+)等于()A.B.C.D.解析:由于cos -cos =-2sinsin=-2sinsin=-sin,因此sin=-,于是cos(+)=1-2sin2=1-2.答案:C7.cos 20+cos 60+cos 100+cos 140的值为.解析:cos 20+cos 60+cos 100+cos 140=cos 20+cos 100+cos 140+=2cos 60cos(-40)+cos 140+=cos 40+cos 140+=cos 40-cos 40+.答案:8.若cos2-cos2=m,则sin(+)sin(-)=.解析:sin(+)sin(-)=- (cos 2-cos 2)=- (2cos2-1)-(2cos2-1)=cos2-cos2=-m.答案:-m9.若x为锐角三角形的内角,则函数y=sin+sin x的值域为.解析:y=2sincossin.由条件,知x+,所以sin1.所以y.答案:10.化简:.解:原式=cot 4.11.已知ABC的三个内角A,B,C满足A+C=2B,=-,求cos的值.解:由题设条件知B=60,A+C=120,-=-=-2,=-2.将上式化简为cos A+cos C=-2cos Acos C,则2coscos=-cos(A+C)+cos(A-C).将cos=cos 60=,cos(A+C)=cos 120=-代入上式,得coscos(A