2017_18學年高中數學第三章3.3三角函數的積化和差與和差化積.docx

3.3三角函數的積化和差與和差化積課時過關能力提升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可化簡為()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(-)=,則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.已知直角三角形中兩銳角為A和B,則sin Asin B()A.有最大值和最小值0B.有最大值,但無最小值C.既無最大值,也無最小值D.有最大值1,但無最小值解析:sin Asin B=-cos(A+B)-cos(A-B)=-cos(A-B).又0A蟺2,0B蟺2,所以-蟺2A-B蟺2,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-=蟺3,且cos -cos =,則cos(+)等于()A.B.C.D.解析:由于cos -cos =-2sinsin=-2sinsin蟺6=-sin,因此sin=-,于是cos(+)=1-2sin2=1-2.答案:C7.cos 20+cos 60+cos 100+cos 140的值為.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,則sin(+)sin(-)=.解析:sin(+)sin(-)=-(cos 2-cos 2)=-(2cos2-1)-(2cos2-1)=cos2-cos2=-m.答案:-m9.若x為銳角三角形的內角,則函數y=sin+sin x的值域為.解析:y=2sincossin.由條件,知蟺6x+,所以sin1.所以y.答案:10.化簡:.解:原式=cot 4.11.已知ABC的三個內角A,B,C滿足A+C=2B,=-,求cos的值.解:由題設條件知B=60,A+C=120,-=-=-2,=-2.將上式化簡為cos A+cos C=-2cos Acos C,則2coscos=-cos(A+C)+cos(A-C).將cos=cos 60=,cos(A+C)=cos 120=-代入上式,得coscos(A