（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時規(guī)范練20 兩角和與差的正弦、余弦與正切公式 文.docx

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課時規(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0yxbcB.bacC.cabD.acb導(dǎo)學(xué)號2419090017.(2017江西重點中學(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因為,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因為tan =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.B0yx2,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時取等號,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,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(-)=-130,-2-0,cos sin 12sin 11,acb.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.