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cosx=,则cos2x=()A.-B.C.-D.2.(2017云南昆明一中仿真,理2)cos70°sin50°-cos200°sin40°的值为()A.-B.-C.D.3.已知α∈,且cosα=-,则tan等于()A.7B.C.-D.-74.设sin,则sin2θ=()A.-B.-C.D.5.若tanα=2tan,则=()A.1B.2C.3D.46.已知cos+sinα=,则sin的值为()A.B.C.-D.-7.若0b>cB.b>a>cC.c>a>bD.a>c>b导学号〚21500533〛17.已知sin,θ∈,则cos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.Dcos2x=2cos2x-1=2-1=2.Dcos70°sin50°-cos200°sin40°=cos70°sin50°+cos20°sin40°=cos70°sin50°+sin70°cos50°=sin(50°+70°)=sin120°=3.B因为,且cosα=-,所以sinα=-,所以tanα=所以tan4.Asin2θ=-cos=2sin2-1=2-1=-5.C因为tanα=2tan,所以=====3.6.C∵cos+sinα=cosα+sinα=,cosα+sinα=∴sin=-sin=-=-7.B∵00,cosα<0.∵3cos2α=sin,∴3(cos2α-sin2α)=(cosα-sinα),∴cosα+sinα=,∴两边平方,可得1+2sinαcosα=,∴sin2α=2sinαcosα=-14.B∵3sin2θ=4tanθ,=4tanθ.∵θ≠kπ(k∈Z),tanθ≠0,=2,解得tan2θ=,∴cos2θ=cos2θ-sin2θ=故选B.15.2令f(x)=4sinx-2sinx-|ln(x+1)|=sin2x-|ln(x+1)|=0,即sin2x=|ln(x+1)|,在同一平面直角坐标系中作出y=sin2x与y=|ln(x+1)|的图象.由图象知共有2个交点,故f(x)的零点个数为2.16.Da=sin40°cos127°+cos40°sin127°=sin(40°+127°)=sin167°=sin13°,b=(sin56°-cos56°)=sin56°-cos56°=sin(56°-45°)=sin11°,c==cos239°-sin239°=cos78°=sin12°,∵sin13°>sin12°>sin11°,∴a>c>b.故选D.17.-由得θ+,又sin,所以cos=-cos=cos=coscos-sinsin=-=-
