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版必修4.doc

3.1.1两角差的余弦公式1.cos(-40)cos 20-sin(-40)sin(-20)等于()a.-12b.-32c.12d.32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=12.答案:c2.计算cos 4-sin+cos的值是()a.2b.-2c.22d.-22解析:原式=cos 4cos+sin 4sinsin+cos=22(cos+sin)sin+cos=22.答案:c3.若a=(cos 60,sin 60),b=(cos 15,sin 15),则ab=()a.22b.12c.32d.-12解析:ab=cos 60cos 15+sin 60sin 15=cos(60-15)=cos 45=22.答案:a4.若sin =35,2,则2cos 4-的值为()a.-25b.-210c.-7210d.-725解析:sin =35,2,cos =-1-sin2=-45,2cos -4=2coscos 4+sinsin 4=-24522+23522=-25.故选a.答案:a5.若sin (+)=-35,是第二象限角,sin 2+=-255,是第三象限角,则cos (-)的值是()a.-55b.55c.11525d.5解析:由sin (+)=-35,得sin =35,又由sin 2+=-255,得cos =-255.是第二象限角,是第三象限角,cos =-45,sin =-55,cos (-)=cos cos +sin sin =-45-255+35-55=55.答案:b6.若cos (-)=55,cos 2=1010,均为锐角,且,则+的值为()a.6b.4c.34d.56解析:,0,2,且,-2-0,sin (-)=-1-cos2(-)=-255.又0,2,2(0,),sin 2=1-cos22=31010.cos (+)=cos 2-(-)=cos 2cos (-)+sin 2sin (-)=101055+31010-255=-22.又+(0,),+=34.答案:c7.已知cos-3=cos ,则tan =.解析:cos-3=cos cos3+sin sin3=12cos +32sin =cos ,32sin =12cos .sincos=33,即tan =33.答案:338.若cos cos -sin sin =15,cos (-)=35,则tan tan =.解析:cos (-)=cos cos +sin sin =35,又cos cos -sin sin =15,+,得cos cos =25,-,得sin sin =15,得tan tan =12.答案:129.导学号08720082已知cos(+30)=513,3090,则cos =.解析:3090,60+30120.又cos(+30)=513,sin(+30)=1-12132=1213.cos =cos (+30)-30=cos(+30)cos 30+sin(+30)sin 30=51332+121312=53+1226.答案:53+122610.已知tan =43,cos(+)=-1114,均为锐角,求cos 的值.解:0,2,tan =43,sin =43cos ,sin2+cos2=1.由得sin =437,cos =17.+(0,),cos(+)=-1114,sin(+)=5314.cos =cos (+)-=cos(+)cos +sin(+)sin =-111417+5314437=12.cos =12.11.若x2,且sin x=45,求2cos x-23+2cos x的值.解:x2,sin x=45,cos x=-35,2cos x-23+2cos x=2cosxcos 23+sinxsin 23+2cos x=2-12cosx+32sinx+2cos x=3sin x+cos x=435-35=43-35.12.导学号08720083已知,34,sin(+)=-35,sin-4=1213,求cos+4的值.解:,34,+32,2,-42,34.又sin(+)=-35,sin-4=1213,cos(+)=1