（福建专用）高考数学总复习 课时规范练20 两角和与差的正弦、余弦与正切公式 文 新人教A版.doc

课时规范练20两角和与差的正弦、余弦与正切公式基础巩固组1.(2017山东,文4)已知cos x=34,则cos 2x=() a.-14b.14c.-18d.182.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2419090017.(2017江西重点中学盟校二模,文14)已知sin+4=14,-32,-,则cos+712的值为.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+