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=34,则cos2x=( )A.-14B.14C.-18D.182.(2022云南昆明一中仿真,理2)cos70°sin50°-cos200°sin40°的值为( )A.-32B.-12C.12D.323.已知α∈π,3π2,且cosα=-45,则tanπ4-α等于( )A.7B.17C.-17D.-74.设sinπ4+θ=13,则sin2θ=( )A.-79B.-19C.19D.795.若tanα=2tanπ5,则cosα-3π10sinα-π5=( )A.1B.2C.3D.46.已知cosα-π6+sinα=435,则sinα+7π6的值为( )A.12B.32C.-45D.-127.若0<y≤x<π2,且tanx=3tany,则x-y的最大值为( )A.π4B.π6C.π3D.π2〚导学号21500531〛8.已知sin2α=2-2cos2α,则tanα= . 9.函数f(x)=sin2xsinπ6-cos2xcos5π6在-π2,π2上的单调递增区间为 . 10.在△ABC中,C=60°,tanA2+tanB2=1,则tanA2tanB2=.11.已知α,β均为锐角,且sinα=35,tan(α-β)=-13.(1)求sin(α-β)的值;(2)求cosβ的值.5\n二、综合提升组12.(2022广西名校联考,理9)已知△ABC的面积为S,且AB·AC=S,则tan2A的值为( )A.12B.2C.34D.-4313.(2022河北衡水中学三调,理3)若α∈π2,π,且3cos2α=sinπ4-α,则sin2α的值为( )A.118B.-118C.1718D.-171814.(2022河北邯郸二模,理5)已知3sin2θ=4tanθ,且θ≠kπ(k∈Z),则cos2θ等于( )A.-13B.13C.-14D.14〚导学号21500532〛15.函数f(x)=4cos2x2cosπ2-x-2sinx-|ln(x+1)|的零点个数为 . 三、创新应用组16.设a=cos50°cos127°+cos40°cos37°,b=22(sin56°-cos56°),c=1-tan239°1+tan239°,则a,b,c的大小关系是( )A.a>b>cB.b>a>cC.c>a>bD.a>c>b〚导学号21500533〛17.已知sinθ+π4=14,θ∈-3π2,-π,则cosθ+7π12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.D cos2x=2cos2x-1=2×342-1=18.2.D cos70°sin50°-cos200°sin40°=cos70°sin50°+cos20°sin40°=cos70°sin50°+sin70°cos50°=sin(50°+70°)=sin120°=32.3.B 因为α∈π,3π2,且cosα=-45,5\n所以sinα=-35,所以tanα=34.所以tanπ4-α=1-tanα1+tanα=1-341+34=17.4.A sin2θ=-cosπ2+2θ=2sin2π4+θ-1=2×132-1=-79.5.C 因为tanα=2tanπ5,所以cosα-3π10sinα-π5=sinα-3π10+π2sinα-π5=sinα+π5sinα-π5=sinαcosπ5+cosαsinπ5sinαcosπ5-cosαsinπ5=tanα+tanπ5tanα-tanπ5=3tanπ5tanπ5=3.6.C ∵cosα-π6+sinα=32cosα+32sinα=435,∴12cosα+32sinα=45.∴sinα+7π6=-sinα+π6=-32sinα+12cosα=-45.7.B ∵0<y≤x<π2,且tanx=3tany,x-y∈0,π2,∴tan(x-y)=tanx-tany1+tanxtany=2tany1+3tan2y=21tany+3tany≤33=tanπ6,当且仅当3tan2y=1时取等号,∴x-y的最大值为π6,故选B.8.0或12 ∵已知sin2α=2-2cos2α=2-2(1-2sin2α)=4sin2α,∴2sinαcosα=4sin2α,∴sinα=0,或cosα=2sinα,即tanα=0,或tanα=12.9.-5π12,π12 f(x)=sin2xsinπ6-cos2xcos5π6=sin2xsinπ6+cos2xcosπ6=cos2x-π6.当2kπ-π≤2x-π6≤2kπ(k∈Z),即kπ-5π12≤x≤kπ+π12(k∈Z)时,函数f(x)单调递增.取k=0,得-5π12≤x≤π12,故函数f(x)在-π2,π2上的单调递增区间为-5π12,π12.10.1-33 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)∵α,β∈0,π2,5\n∴-π2<α-β<π2.又tan(α-β)=-13<0,∴-π2<α-β<0.由tan(α-β)=-13,sin2(α-β)+cos2(α-β)=1,解得sin(α-β)=-1010.(2)由(1)可得,cos(α-β)=31010.∵α为锐角,且sinα=35,∴cosα=45.∴cosβ=cos[α-(α-β)]=cosαcos(α-β)+sinαsin(α-β)=45×31010+35×-1010=91050.12.D 设△ABC的角A,B,C所对应的边分别为a,b,c.∵AB·AC=S,∴bccosA=12bcsinA,∴tanA=2,∴tan2A=2tanA1-tan2A=2×21-22=-43,故选D.13.D ∵α∈π2,π,∴sinα>0,cosα<0.∵3cos2α=sinπ4-α,∴3(cos2α-sin2α)=22(cosα-sinα),∴cosα+sinα=26,∴两边平方,可得1+2sinαcosα=118,∴sin2α=2sinαcosα=-1718.14.B ∵3sin2θ=4tanθ,∴6sinθcosθsin2θ+cos2θ=6tanθ1+tan2θ=4tanθ.∵θ≠kπ(k∈Z),tanθ≠0,∴31+tan2θ=2,解得tan2θ=12,∴cos2θ=cos2θ-sin2θ=cos2θ-sin2θcos2θ+sin2θ=1-tan2θ1+tan2θ=1-121+12=13.故选B.15.2 令f(x)=4·1+cosx2·sinx-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.D a=sin40°cos127°+cos40°sin127°=sin(40°+127°)=sin167°=sin13°,b=22(sin56°-cos56°)=22sin56°-22cos56°=sin(56°-45°)=sin11°,c=1-tan239°1+tan239°=cos239°-sin239°cos239°cos239°+sin239°cos239°=cos239°-sin239°=cos78°=sin12°,5\n∵sin13°>sin12°>sin11°,∴a>c>b.故选D.17.-15+38 由θ∈-3π2,-π得θ+π4∈-5π4,-3π4,又sinθ+π4=14,所以cosθ+π4=-154.cosθ+7π12=cosθ+π4+π3=cosθ+π4cosπ3-sinθ+π4sinπ3=-154×12-14×32=-15+38.5
