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 =,则sin 2等于()A.-B.-C.D.解析:由已知得cos =-=-,于是sin 2=2sin cos =2=-.答案:A2.等于()A.-sin 50B.sin 50C.-cos 50D.cos 50解析:cos 50.答案:D3.已知向量a=(3,-2),b=(cos ,sin ),若ab,则tan 2的值为()A.B.-C.D.-解析:由ab得3sin =-2cos ,于是tan =-,从而tan 2=-.答案:B4.已知sin,则sin 2等于()A.-B.C.-D.解析:由已知得sin cos+cos sin,于是(sin +cos )=,sin +cos =,从而(sin +cos )2=,即1+sin 2=,故sin 2=-.答案:C5.函数y=2sin x(sin x+cos x)的最大值为()A.1+B.-1C.D.2解析:y=2sin x(sin x+cos x)=2sin2x+2sin xcos x=1-cos 2x+sin 2x=sin+1,因此当sin=1时,函数取最大值+1.答案:A6.已知,则tan +=()A.-8B.8C.D.-解析:=cos -sin =,1-2sin cos =,即sin cos =-.则tan +=-8.故选A.答案:A7.已知sin =,则sin=.解析:sin=sin=-cos 2=-(1-2sin2)=2-1=2-.答案:2-8.sin 10sin 30sin 50sin 70的值等于.解析:sin 10sin 50sin 70=.故sin 10sin 30sin 50sin 70=.答案:9.已知=-5,则3cos 2+sin 2=.解析:由=-5,得2sin +cos =-5sin +15cos ,7sin =14cos .tan =2.3cos 2+sin 2=3(cos2-sin2)+2sin cos =3=-1.答案:-110.已知为锐角,且sin =.(1)求的值;(2)求tan的值.解:(1)为锐角,且sin =,cos =.=20.(2)由(1),得tan =,故tan.11.已知向量m=(sin x,-1),向量n=,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)在上的最大值,求锐角A.解:(1)f(x)=(m+n)m=sin2x+sin xcos x+sin 2x+sin 2x-cos 2x+2=sin+2,所以函数f(x)的最小正周期T=.(2)由(1),知f(x)=sin+2.当x时,-2x-.由正弦函数的图象可知,当2x-时,f(x)取得最大值3,即f(A)=3,此时2A-,所以A=.