1、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=,则cos 2x=()A.-B.C.-D.2.cos 70sin 50-cos 200sin 40的值为()A.-B.-C.D.3.已知,且cos =-,则tan等于()A.7B.C.-D.-74.设sin,则sin 2=()A.-B.-C.D.5.若tan =2tan,则=()A.1B.2C.3D.46.已知cos+sin =,则sin的值为()A.B.C.-D.-7.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,则c
2、os的值为.课时规范练20两角和与差的正弦、余弦与正切公式1.Dcos 2x=2cos2x-1=2-1=.2.Dcos 70sin 50-cos 200sin 40=cos 70sin 50+cos 20sin 40=cos 70sin 50+sin 70cos 50=sin(50+70)=sin 120=.3.B因为,且cos =-,所以sin =-,所以tan =.所以tan.4.Asin 2=-cos=2sin2-1=2-1=-.5.C因为tan =2tan,所以=3.6.Ccos+sin =cos +sin =,cos +sin =.sin=-sin=-=-.7.B0yx,且tan x
3、=3tan y,x-y,tan(x-y)=tan,当且仅当3tan2y=1时取等号,x-y的最大值为,故选B.8.0或已知sin 2=2-2cos 2=2-2(1-2sin2)=4sin2,2sin cos =4sin2,sin =0,或cos =2sin ,即tan =0,或tan =.9.f(x)=sin 2xsin-cos 2xcos=sin 2xsin+cos 2xcos=cos.当2k-2x-2k(kZ),即k-xk+(kZ)时,函数f(x)单调递增.取k=0,得-x,故函数f(x)在上的单调递增区间为.10.1-由C=60,则A+B=120,即=60.根据tan,又tan+tan=1,得,解得tantan=1-.11.解 (1),-.又tan(-)=-0,-0,cos sin 12sin 11,acb.故选D.17.-由得+,又sin,所以cos=-.cos=cos=coscos-sinsin=-=-.
