高中数学第三章三角恒等变换3.1.1两角差的余弦公式课后习题新人教A版.docx

3.1.1两角差的余弦公式1.cos(-40)cos 20-sin(-40)sin(-20)等于()A.-B.-C.D.解析:cos(-40)cos 20-sin(-40)sin(-20)=cos(-40)cos 20+sin(-40)sin 20=cos(-40-20)=cos(-60)=cos 60=.答案:C2.计算的值是()A.B.-C.D.-解析:原式=.答案:C3.若a=(cos 60,sin 60),b=(cos 15,sin 15),则ab=()A.B.C.D.-解析:ab=cos 60cos 15+sin 60sin 15=cos(60-15)=cos 45=.答案:A4.若sin =,则2cos 的值为()A.-B.-C.-D.-解析:sin =,cos =-=-,2cos =2=-2+2=-.故选A.答案:A5.若sin (+)=-,是第二象限角,sin =-,是第三象限角,则cos (-)的值是()A.-B.C.D.解析:由sin (+)=-,得sin =,又由sin =-,得cos =-.是第二象限角,是第三象限角,cos =-,sin =-,cos (-)=cos cos +sin sin =.答案:B6.若cos (-)=,cos 2=,均为锐角,且,则+的值为()A.B.C.D.解析:,且,-0,sin (-)=-=-.又,2(0,),sin 2=.cos (+)=cos 2-(-)=cos 2cos (-)+sin 2sin (-)=-.又+(0,),+=.答案:C7.已知cos=cos ,则tan =.解析:cos=cos cos+sin sin=cos +sin =cos ,sin =cos .,即tan =.答案:8.若cos cos -sin sin =,cos (-)=,则tan tan =.解析:cos (-)=cos cos +sin sin =,又cos cos -sin sin =,+,得cos cos =,-,得sin sin =,得tan tan =.答案:9.导学号08720082已知cos(+30)=,3090,则cos =.解析:3090,60+30120.又cos(+30)=,sin(+30)=.cos =cos (+30)-30=cos(+30)cos 30+sin(+30)sin 30=.答案:10.已知tan =4,cos(+)=-,均为锐角,求cos 的值.解:,tan =4,sin =4cos ,sin2+cos2=1.由得sin =,cos =.+(0,),cos(+)=-,sin(+)=.cos =cos (+)-=cos(+)cos +sin(+)sin =.cos =.11.若x,且sin x=,求2cos +2cos x的值.解:x,sin x=,cos x=-,2cos +2cos x=2+2cos x=2+2cos x=sin x+cos x=.12.导学号08720083已知,sin(+)=-,sin,求cos的值.