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课时规范练21两角和与差的正弦、余弦与正切公式一、基础巩固组1.已知cos x=34,则cos 2x=()A.-14B.14C.-18D.182.(2017云南昆明一中仿真,理2)cos 70sin 50-cos 200sin 40的值为()A.-32B.-12C.12D.323.已知,32,且cos =-45,则tan4-等于()A.7B.17C.-17D.-74.设sin4+=13,则sin 2=()A.-79B.-19C.19D.795.若tan =2tan5,则cos-310sin-5=()A.1B.2C.3D.46.已知cos-6+sin =435,则sin+76的值为()A.12B.32C.-45D.-127.若0yxbcB.bacC.cabD.acb导学号2150053317.已知sin+4=14,-32,-,则cos+712的值为.课时规范练21两角和与差的正弦、余弦与正切公式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因为,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因为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,当且仅当3tan2y=1时取等号,x-y的最大值为6,故选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.当2k-2x-62k(kZ),即k-512xk+12(kZ)时,函数f(x)单调递增.取k=0,得-512x12,故函数f(x)在-2,2上的单调递增区间为-512,12.10.1-33由C=60,则A+B=120,即A2+B2=60.根据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.故选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.
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