1、3.1.3二倍角的正弦、余弦、正切公式课时过关能力提升基础巩固1.12-sin215的值是()A.64B.6-24C.32D.34解析:原式=12-1-cos(215)2=cos302=34.答案:D2.已知sin -cos =2,(0,),则sin 2=()A.-1B.-22C.22D.1解析:将sin -cos =2两端同时平方得,(sin -cos )2=2,整理得1-2sin cos =2,于是sin 2=2sin cos =-1,故选A.答案:A3.若tan +1tan=4,则sin 2=()A.15B.14C.13D.12解析:由tan +1tan=4,得sincos+cossin
2、=4,即2sin2=4,sin 2=12.答案:D4.若x=12,则cos2x-sin2x的值等于()A.14B.12C.22D.32解析:当x=12时,cos2x-sin2x=cos 2x=cos212=cos6=32.答案:D5.化简1-sin1+1-cos12的结果是()A.sin12B.cos12C.2sin12-cos12D.2cos12-sin12答案:B6.tan81-tan28=.解析:原式=122tan81-tan28=12tan28=12tan4=12.答案:127.在ABC中,若cos A=513,则sin 2A=,cos 2A=,tan 2A=.解析:0A,sin A=
3、1-cos2A=1213.sin 2A=2sin Acos A=120169,cos 2A=2cos2A-1=25132-1=50169-1=-119169,tan 2A=sin2Acos2A=-120119.答案:120169-119169-1201198.已知cos 2=23,则cos4+sin4=.解析:cos4+sin4=(cos2+sin2)2-2sin2cos2=1-12sin22=1-12(1-cos22)=1-121-232=1118.答案:11189.已知函数f(x)=sin2x+2sin xcos x+3cos2x,xR,求f(x)的周期及值域.解:f(x)=1-cos2x
4、2+sin 2x+3(1+cos2x)2=2+sin 2x+cos 2x=2+2sin2x+4.函数f(x)的周期为,值域为2-2,2+2.10.在ABC中,若sin Asin B=cos2C2,试判断ABC的形状.解:sin Asin B=cos2C2=1+cosC2=1-cos(A+B)2,即2sin Asin B+cos(A+B)=1,2sin Asin B+cos Acos B-sin Asin B=cos Acos B+sin Asin B=cos(A-B)=1.-A-B,A-B=0,即A=B.ABC是等腰三角形.能力提升1.函数y=sin2x+6sin2x+23的最小正周期为 ()
5、A.2B.4C.2D.解析:y=sin2x+6sin2x+6+2=sin2x+6cos2x+6=12sin4x+3.所以该函数的周期为2.答案:A2.已知为第二象限角,sin +cos =33,则cos 2等于 ()A.-53B.-59C.59D.53解析:由题意,得(sin +cos )2=13,1+sin 2=13,sin 2=-23.为第二象限角,cos -sin 0,cos 0,且|cos |sin |,cos 2=cos2-sin20,cos 2=-1-sin22=-1-232=-1-49=-53,故选A.答案:A3.若sin6-=13,则cos23+2的值是()A.-79B.-13
6、C.13D.79解析:cos23+2=cos-3-2=-cos3-2=-1-2sin26-=2sin26-1=2132-1=-79.答案:A4.已知3+tan1-tan=1+23,则sin2+sin 2的值为()A.1B.45C.35D.25解析:由3+tan1-tan=1+23,得tan =12.sin2+sin 2=sin2+2sincossin2+cos2=tan2+2tantan2+1=14+114+1=1.答案:A5.已知tan 2=2,则1-cos+sin1+cos+sin=.解析:1-cos+sin1+cos+sin=2sin22+2sin2cos22cos22+2sin2cos
7、2=2sin2sin2+cos22cos2cos2+sin2=tan2=2.答案:26.sin 10sin 30sin 50sin 70的值等于.解析:sin 10sin 50sin 70=sin20sin50sin702cos10=sin20cos20sin502cos10=sin40sin504cos10=sin40cos404cos10=sin808cos10=18,sin 10sin 30sin 50sin 70=116.答案:1167.已知函数f(x)=sin x-23sin2x2.(1)求f(x)的最小正周期;(2)求f(x)在区间0,23上的最小值.解:(1)因为f(x)=sin
8、 x+3cos x-3=2sinx+3-3,所以f(x)的最小正周期为2.(2)因为0x23,所以3x+3.当x+3=,即x=23时,f(x)取得最小值.所以f(x)在区间0,23上的最小值为f23=-3.8.设向量a=(3sin x,sin x),b=(cos x,sin x),x0,2.(1)若|a|=|b|,求x的值;(2)设函数f(x)=ab,求f(x)的最大值.解:(1)由|a|2=(3sin x)2+sin2x=4sin2x,|b|2=cos2x+sin2x=1,及|a|=|b|,得4sin2x=1.又x0,2,从而sin x=12,所以x=6.(2)f(x)=ab=3sin xcos x+sin2x=32sin 2x-12cos 2x+12=sin2x-6+12,当x=30,2时,sin2x-6取最大值1.所以f(x)的最大值为32.