3.1.3　二倍角的正弦、余弦、正切公式

3.1.3二倍角的正弦、余弦、正切公式【選題明細(xì)表】 知識(shí)點(diǎn)、方法題號(hào)化簡(jiǎn)求值1,2,5,7條件求值4,6,9,10,12綜合應(yīng)用3,8,11,131.化簡(jiǎn)1-2cos2(4-)的結(jié)果為(D)(A)2cos 2(B)-cos 2(C)sin 2(D)-sin 2解析:1-2cos2(4-)=1-1+cos2(4-)=-cos(2-2)=-sin 2.故選D.2.(2019玉溪模擬)下列各式中,值為12的是(D)(A)sin 15cos15(B)cos212-sin212(C)1+cos 62 (D)tan22.51-tan222.5解析:sin 15cos 15=12sin 30=14,排除A項(xiàng).cos212-sin2 12=cos 6=32,排除B項(xiàng).1+cos 62=1+322=2+32,排除C項(xiàng).由tan 45=2tan22.51-tan222.5=1,知選D.3.(2019東莞模擬)已知函數(shù)f(x)=cos2x-1sin2x,則有(B)(A)函數(shù)f(x)的圖象關(guān)于直線x=2對(duì)稱(B)函數(shù)f(x)的圖象關(guān)于點(diǎn)(2,0)對(duì)稱(C)函數(shù)f(x)的最小正周期為2(D)函數(shù)f(x)在區(qū)間(0,)內(nèi)單調(diào)遞減解析:因?yàn)閒(x)=cos2x-1sin2x=-2sin2x2sinxcosx=-tan x,所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(2,0)對(duì)稱,故選B.4.(2019綿陽(yáng)模擬)若tan(-4)=2,則tan 2等于(D)(A)-3(B)3(C)-34(D)34解析:因?yàn)閠an (-4)=2=tan-11+tan,可求得tan =-3,所以tan 2=2tan1-tan2=2(-3)1-(-3)2=34.故選D.5.若f(x)=2tan x-2sin2x2-1sin x2cos x2,則f(12)的值為(B)(A)-433 (B)8(C)43 (D)-43解析:f(x)=2sinxcosx-2cosxsinx=2sin2x+cos2xsinxcosx=4sin2x,所以f(12)=4sin 6=8.故選B.6.已知為第二象限角,sin =35,則tan 2=.解析:由sin =35,且為第二象限角得,cos =-1-sin2=-45,得tan =sincos=-34,tan 2=2tan1-tan2=-247.答案:-2477.若0sin20,所以原式=sin2+cos2-cos2+sin2=2sin2.答案:2sin28.(2019安徽合肥高一檢測(cè))求證:1cos2-tan tan 2=1.證明:1cos2-tan tan 2=1cos2-sinsin2coscos2=cos-2sin2coscoscos2=1-2sin2cos2=cos2cos2=1.9.(2019榆林二模)已知cossin=3cos(2+),|2,則sin 2 等于(C)(A)829 (B)223 (C)429 (D)229解析:因?yàn)閨2,所以cos (0,1,因?yàn)閏ossin=3cos(2+),所以cossin=3cos ,可得sin =13,所以cos =1-sin2=223,所以sin 2=2sin cos =213223=429.故選C.10.(2019云南昆明模擬)已知sin 2=35(42),tan(-)=12,則tan(+)等于(A)(A)-2 (B)-1 (C)-1011 (D)-211解析:因?yàn)?2,所以22.因?yàn)閟in 2=35,所以cos 2=-45,tan 2=-34,所以tan(+)=tan2-(-)=tan2-tan(-)1+tan2tan(-)=-34-121-3412=-2.故選A.11.(2019南陽(yáng)期末)已知:sin +cos =32,則cos 2+cos 2的取值范圍是.解析:因?yàn)閟in +cos =32,所以cos 2+cos 2=1-2sin2+2cos2-1=2(sin +cos )(cos -sin )=3(cos -sin ),由sin +cos =32得,cos =32-sin ,易得sin 12,1,所以cos -sin =32-2sin -12,12,所以cos 2+cos 2-32,32.答案:-32,3212.(2019安徽合肥高一檢測(cè))已知cos =-34,sin =23,是第三象限角,(2,).(1)求sin 2的值;(2)求cos(2+)的值.解:(1)因?yàn)槭堑谌笙藿?cos =-34,所以sin =-1-cos2=-74,所以sin 2=2sin cos =2(-74)(-34)=378.(2)因?yàn)?2,),sin =23,所以cos =-1-sin2=-53,因?yàn)閏os =-34,所以cos 2=2cos2-1=2916-1=18,所以cos(2+)=cos 2cos -sin 2sin =18(-53)-37823=-5+6724.13.設(shè)函數(shù)f(x)=53cos2x+3sin2x-4sin xcos x.(1)求f(512);(2)若f()=53,(2,),求角.解:f(x)=53cos2x+3sin2x-4sin x cos x=53cos2x+53sin2x-2sin 2x-43sin2x=53-2sin 2x-23(1-cos 2x)=33-2sin 2x+23cos 2x=33-4(sin 2x12-cos 2x32)=33-4(sin 2x cos3-cos 2x sin