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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)=-12-12-0=14.答案:A2.式子sin 20+cos 10可化简为()A.sin 50B.cos 50C.3sin 50D.3cos 50解析:sin 20+cos 10=sin 20+sin 80=2sin 50cos 30=3sin 50.答案:C3.若cos(+)cos(-)=,则cos2-sin2等于()A.-B.-C.D.23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=13.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)=-12cos2-cos(A-B)=12cos(A-B).又0A2,0B2,所以-2A-B2,于是0cos(A-B)1,故0sin Asin B.答案:B5.cos 72-cos 36等于()A.3-23B.C.-D.-解析:cos 72-cos 36=-2sin 54sin 18=-2sin 18cos 36=-2sin18cos18cos36cos18=-sin36cos36cos18=-2sin36cos362cos18=-sin722cos18=-sin722sin72=-12.答案:C6.已知-=3,且cos -cos =,则cos(+)等于()A.B.C.D.解析:由于cos -cos =-2sin+2sin-2=-2sin+2sin6=-sin+2=13,因此sin+2=-,于是cos(+)=1-2sin2+2=1-2-132=79.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+12=2cos 60cos(-40)+cos 140+12=cos 40+cos 140+12=cos 40-cos 40+12=12.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=sinx+3+sin x的值域为.解析:y=2sinx+6cos6=3sinx+6.由条件,知6x+623,所以12sinx+61.所以y32,3.答案:32,310.化简:cos+cos3+cos5+cos7sin+sin3+sin5+sin7.解:原式=(cos+cos7)+(cos3+cos5)(sin+sin7)+(sin3+sin5)=2cos4cos3+2cos4cos2sin4cos3+2sin4cos=2cos4(cos3+cos)2sin4(cos3+cos)=cot 4.11.已知ABC的三个内角A,B,C满足A+C=2B,1cosA+1cosC=-2cosB,求cosA-C2的值.解:由题设条件知B=60,A+C=120,-2cosB=-2cos60=-22,1cosA+1cosC=-22.将上式化简为cos A+cos C=-22cos Acos C,则2cosA+C2cosA-C2=-2cos(A+C)+cos(A-C).将cosA+C2=cos 60=12,cos(A+C)=cos 120=-12代入上式,得cosA-C2=22-2cos(A-C).将cos(A-C)=2cos2A-C2-1代入上式并整理,得42cos2A-C2+2cosA-C2-32=0,即2cosA-C2-222cosA-C2+3=0.22cosA-C2+30,2cosA-C2-2=0.cosA-C2=22.
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