1、3.2倍角公式和半角公式3.2.1倍角公式课时过关能力提升1.已知为第二象限的角,sin =35,则sin 2等于()A.-2425B.-1225C.1225D.2425解析:由已知得cos =-1-sin2=-45,于是sin 2=2sin cos =235-45=-2425.答案:A2.1+cos100等于()A.-2sin 50B.2sin 50C.-2cos 50D.2cos 50解析:1+cos100=1+2cos250-1=2cos250=2cos 50.答案:D3.已知向量a=(3,-2),b=(cos ,sin ),若ab,则tan 2的值为()A.125B.-125C.121
2、3D.-1213解析:由ab得3sin =-2cos ,于是tan =-23,从而tan 2=2tan1-tan2=-431-49=-125.答案:B4.已知sin+4=55,则sin 2等于()A.-45B.45C.-35D.35解析:由已知得sin cos4+cos sin4=55,于是22(sin +cos )=55,sin +cos =105,从而(sin +cos )2=25,即1+sin 2=25,故sin 2=-35.答案:C5.函数y=2sin x(sin x+cos x)的最大值为()A.1+2B.2-1C.2D.2解析:y=2sin x(sin x+cos x)=2sin2
3、x+2sin xcos x=1-cos 2x+sin 2x=2sin2x-4+1,因此当sin2x-4=1时,函数取最大值2+1.答案:A6.已知cos22sin+4=52,则tan +1tan=()A.-8B.8C.18D.-18解析:cos22sin+4=cos2-sin2222sin+22cos=cos -sin =52,1-2sin cos =54,即sin cos =-18.则tan +1tan=sincos+cossin=sin2+cos2sincos=1-18=-8.故选A.答案:A7.已知sin =5-12,则sin2-4=.解析:sin2-4=sin2-2=-cos 2=-(
4、1-2sin2)=25-122-1=2-5.答案:2-58.sin 10sin 30sin 50sin 70的值等于.解析:sin 10sin 50sin 70=sin20sin50sin702cos10=sin20cos20sin502cos10=sin40sin504cos10=sin40cos404cos10=sin808cos10=18.故sin 10sin 30sin 50sin 70=116.答案:1169.已知2sin+cossin-3cos=-5,则3cos 2+sin 2=.解析:由2sin+cossin-3cos=-5,得2sin +cos =-5sin +15cos ,7
5、sin =14cos .tan =2.3cos 2+sin 2=3(cos2-sin2)+2sin cos =3(cos2-sin2)cos2+sin2+2sincossin2+cos2=31-tan21+tan2+2tan1+tan2=3-3tan2+2tan1+tan2=-1.答案:-110.已知为锐角,且sin =45.(1)求sin2+sin2cos2+cos2的值;(2)求tan-54的值.解:(1)为锐角,且sin =45,cos =1-sin2=35.sin2+sin2cos2+cos2=sin2+2sincos3cos2-1=452+245353352-1=20.(2)由(1)
6、,得tan =sincos=43,故tan-54=tan-tan541+tantan54=tan-11+tan=17.11.已知向量m=(sin x,-1),向量n=3cosx,-12,函数f(x)=(m+n)m.(1)求f(x)的最小正周期T;(2)已知f(A)恰是f(x)在0,2上的最大值,求锐角A.解:(1)f(x)=(m+n)m=sin2x+3sin xcos x+32=1-cos2x2+32sin 2x+32=32sin 2x-12cos 2x+2=sin2x-6+2,所以函数f(x)的最小正周期T=22=.(2)由(1),知f(x)=sin2x-6+2.当x0,2时,-62x-656.由正弦函数的图象可知,当2x-6=2时,f(x)取得最大值3,即f(A)=3,此时2A-6=2,所以A=3.