2019-2020学年新一线同步人教A版数学必修一练习：5.5.1　第3课时　二倍角的正弦、余弦、正切公式 含解析.docx

教学资料范本2019-2020学年新一线同步人教A版数学必修一练习：5.5.1第3课时二倍角的正弦、余弦、正切公式 含解析编 辑：_时 间：_3课时二倍角的正弦、余弦、正切公式课后篇巩固提升基础巩固1.=()A.-B.-C.D.解析原式=cos2-sin2=cos,故选D.答案D2.若tan =3,则的值等于()A.2B.3C.4D.6解析=2tan =23=6.答案D3.已知sin,则cos的值为()A.B.C.D.解析cos=cos =1-2sin2=1-2.答案D4.若为锐角,3sin =tan =tan ,则tan 2等于()A.B.C.-D.-解析因为为锐角,3sin =tan ,所以cos =,则tan =2,即tan =2,所以tan 2=-.答案D5.若,则tan 2=()A.-B.C.-D.解析等式左边分子、分母同时除以cos (显然cos 0),得,解得tan =-3,tan 2=.答案B6.已知,sin =,则tan 2=.解析由,sin =,得cos =-,tan =-,tan 2=-.答案-7.化简:=.解析原式=tan 2.答案tan 28.若cos(75+)=,则sin(60+2)=.解析依题意,cos(75+)=,则cos(150+2)=2cos2(+75)-1=2-1=-,sin(60+2)=-cos(90+60+2)=-cos(150+2)=.答案9.求下列各式的值:(1);(2)2tan 15+tan215;(3)sin 10sin 30sin 50sin 70.解(1)原式=1.(2)原式=tan 30(1-tan215)+tan215=(1-tan215)+tan215=1.(3)(方法一)sin 10sin 30sin 50sin 70=cos 20cos 40cos 80=.(方法二)令x=sin 10sin 50sin 70,y=cos 10cos 50cos 70.则xy=sin 10cos 10sin 50cos 50sin 70cos 70=sin 20sin 100sin 140=sin 20sin 80sin 40=cos 10cos 50cos 70=y.y0,x=.从而有sin 10sin 30sin 50sin 70=.10.已知sin +cos =,sin,.(1)求sin 2和tan 2的值;(2)求cos(+2)的值.解(1)由题意得(sin +cos )2=,即1+sin 2=,sin 2=,又易知2,cos 2=,tan 2=.(2),-,sin,cos,sin 2=2sincos.又sin 2=-cos 2,cos 2=-.又易知2,sin 2=.又cos2=,cos =,sin =,cos(+2)=cos cos 2-sin sin 2=-.能力提升1.4sin 80-=()A.B.-C.D.2-3解析4sin 80-=-.答案B2.若,且cos2+cos,则tan =()A.B.C.D.或-7解析cos2+cos=cos2-sin 2=cos2-2sin cos =,整理得3tan2+20tan -7=0,解得tan =或tan =-7.又,所以tan =,故选C.答案C3.设sin 2=-sin ,则tan 2的值是.解析sin 2=2sin cos =-sin ,cos =-,又,sin =,tan =-,tan 2=.答案4.已知角,为锐角,且1-cos 2=sin cos ,tan(-)=,则=.解析由1-cos 2=sin cos ,得1-(1-2sin2)=sin cos ,即2sin2=sin cos .为锐角,sin 0