課時(shí)規(guī)范練20兩角和與差的正弦、余弦與正切公式基礎(chǔ)鞏固組1.(2017山東,文4)已知cos x=34,則cos 2x=() A.-14B.14C.-18D.182.cos 70sin 50-cos 200sin 40的值為()A.-32B.-12C.12D.323.已知,32,且cos =-45,則tan4-等于()A.7B.17C.-17D.-74.設(shè)sin4+=13,則sin 2=()A.-79B.-19C.19D.795.若tan =2tan5,則cos-310sin-5=()A.1B.2C.3D.46.已知cos-6+sin =435,則sin+76的值為()A.12B.32C.-45D.-127.若0<yx<2,且tan x=3tan y,則x-y的最大值為()A.4B.6C.3D.2導(dǎo)學(xué)號(hào)241908968.(2017安徽合肥一模,文15)已知sin 2=2-2cos 2,則tan =.9.函數(shù)f(x)=sin 2xsin6-cos 2xcos56在-2,2上的單調(diào)遞增區(qū)間為.10.(2017湖北武漢二月調(diào)考,文14)在ABC中,C=60,tanA2+tanB2=1,則tanA2tanB2=.11.已知,均為銳角,且sin =35,tan(-)=-13.(1)求sin(-)的值;(2)求cos 的值.導(dǎo)學(xué)號(hào)24190897綜合提升組12.已知ABC的面積為S,且ABAC=S,則tan 2A的值為()A.12B.2C.34D.-4313.若2,且3cos 2=sin4-,則sin 2的值為()A.118B.-118C.1718D.-171814.(2017河北邯鄲二模,文5)已知3sin 2=4tan ,且k(kZ),則cos 2等于()A.-13B.13C.-14D.14導(dǎo)學(xué)號(hào)2419089815.函數(shù)f(x)=4cos2x2cos2-x-2sin x-|ln(x+1)|的零點(diǎn)個(gè)數(shù)為.導(dǎo)學(xué)號(hào)24190899創(chuàng)新應(yīng)用組16.(2017河南洛陽(yáng)一模,文5)設(shè)a=cos 50cos 127+cos 40cos 37,b=22(sin 56-cos 56),c=1-tan2391+tan239,則a,b,c的大小關(guān)系是()A.a>b>cB.b>a>cC.c>a>bD.a>c>b導(dǎo)學(xué)號(hào)2419090017.(2017江西重點(diǎn)中學(xué)盟校二模,文14)已知sin+4=14,-32,-,則cos+712的值為.答案：1.Dcos 2x=2cos2x-1=2342-1=18.2.Dcos 70sin 50-cos 200sin 40=cos 70sin 50+cos 20sin 40=cos 70sin 50+sin 70cos 50=sin(50+70)=sin 120=32.3.B因?yàn)?32,且cos =-45,所以sin =-35,所以tan =34.所以tan4-=1-tan1+tan=1-341+34=17.4.Asin 2=-cos2+2=2sin24+-1=2132-1=-79.5.C因?yàn)閠an =2tan5,所以cos-310sin-5=sin-310+2sin-5=sin+5sin-5=sincos5+cossin5sincos5-cossin5=tan+tan5tan-tan5=3tan5tan5=3.6.Ccos-6+sin =32cos +32sin =435,12cos +32sin =45.sin+76=-sin+6=-32sin+12cos=-45.7.B0<yx<2,且tan x=3tan y,x-y0,2,tan(x-y)=tanx-tany1+tanxtany=2tany1+3tan2y=21tany+3tany33=tan6,當(dāng)且僅當(dāng)3tan2y=1時(shí)取等號(hào),x-y的最大值為6,故選B.8.0或12已知sin 2=2-2cos 2=2-2(1-2sin2)=4sin2,2sin cos =4sin2,sin =0,或cos =2sin ,即tan =0,或tan =12.9.-512,12f(x)=sin 2xsin6-cos 2xcos56=sin 2xsin6+cos 2xcos6=cos2x-6.當(dāng)2k-2x-62k(kZ),即k-512xk+12(kZ)時(shí),函數(shù)f(x)單調(diào)遞增.取k=0,得-512x12,故函數(shù)f(x)在-2,2上的單調(diào)遞增區(qū)間為-512,12.10.1-33由C=60,則A+B=120,即A2+B2=60.根據(jù)tanA2+B2=tanA2+tanB21-tanA2tanB2,又tanA2+tanB2=1,得3=11-tanA2tanB2,解得tanA2tanB2=1-33.11.解 (1),0,2,-2<-<2.又tan(-)=-13<0,-2<-<0.由tan(-)=-13,sin2(-)+cos2(-)=1,解得sin(-)=-1010.(2)由(1)可得,cos(-)=31010.為銳角,且sin =35,cos =45.cos =cos -(-)=cos cos(-)+sin sin(-)=4531010+35-1010=91050.12.D設(shè)ABC的角A,B,C所對(duì)應(yīng)的邊分別為a,b,c.ABAC=S,bccos A=12bcsin A,tan A=2,tan 2A=2tanA1-tan2A=221-22=-43,故選D.13.D2,sin >0,cos <0.3cos 2=sin4-,3(cos2-sin2)=22(cos -sin ),cos +sin =26,兩邊平方,可得1+2sin cos =118,sin 2=2sin cos =-1718.14.B3sin 2=4tan ,6sincossin2+cos2=6tan1+tan2=4tan .k(kZ),tan 0,31+tan2=2,解得tan2=12,cos 2=cos2-sin2=cos2-sin2cos2+sin2=1-tan21+tan2=1-121+12=13.故選B.15.2令f(x)=41+cosx2sin x-2sin x-|ln(x+1)|=sin 2x-|ln(x+1)|=0,即sin 2x=|ln(x+1)|,在同一平面直角坐標(biāo)系中作出y=sin 2x與y=|ln(x+1)|的圖象.由圖象知共有2個(gè)交點(diǎn),故f(x)的零點(diǎn)個(gè)數(shù)為2.16.Da=sin 40cos 127+cos 40sin 127=sin(40+127)=sin 167=sin 13,b=22(sin 56-cos 56)=22sin 56-22cos 56=sin(56-45)=sin 11,c=1-tan2391+tan239=cos239-sin239cos239cos239+sin239cos239=cos239-sin239=cos 78=sin 12,sin 13>sin 12>sin 11,a>c>b.故選D.17.-15+38由-32,-得+4-54,-34,又sin+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-15412-1432=-15+38.