課時規(guī)范練21兩角和與差的正弦、余弦與正切公式一、基礎(chǔ)鞏固組1.已知cosx=34,則cos2x=(A.14 B.C.18 D.2.(2017云南昆明一中仿真,理2)cos70°sin50°cos200°sin40°的值為()A.32B.BC.12 D.3.已知α∈π,3π2,且cosα=45,則tanπA.7 B.1C.17D.D4.設(shè)sinπ4+θ=13,則A.79B.BC.19 D.5.若tanα=2tanπ5,則cosα-3A.1 B.2 C.3 D.46.已知cosα-π6+sinα=435,則sinαA.12 B.3C.45D.D7.若0<y≤x<π2,且tanx=3tany,則xy的最大值為()A.π4 B.πC.π3 D.π2?導(dǎo)學(xué)8.已知sin2α=22cos2α,則tanα=.

9.函數(shù)f(x)=sin2xsinπ6cos2xcos5π6在10.在△ABC中,C=60°,tanA2+tanB2=1,則tanA2tanB11.已知α,β均為銳角,且sinα=35,tan(αβ)=1(1)求sin(αβ)的值;(2)求cosβ的值.二、綜合提升組12.(2017廣西名校聯(lián)考,理9)已知△ABC的面積為S,且AB·AC=S,則tan2A的值為(A.12 B.2C.34D.D13.(2017河北衡水中學(xué)三調(diào),理3)若α∈π2,π,且3cos2α=sinπ4-α,則A.118B.B1C.1718D.D14.(2017河北邯鄲二模,理5)已知3sin2θ=4tanθ,且θ≠kπ(k∈Z),則cos2θ等于()A.13 B.1C.14 D.14?導(dǎo)學(xué)15.函數(shù)f(x)=4cos2x2cosπ2-x2sinx|ln(x+1)|三、創(chuàng)新應(yīng)用組16.設(shè)a=cos50°cos127°+cos40°cos37°,b=22(sin56°cos56°),c=1-tan239°1+tan239A.a>b>cB.bB>a>cC.c>a>bD.aD>c>b?導(dǎo)學(xué)號?17.已知sinθ+π4=14,θ∈-3π課時規(guī)范練21兩角和與差的正弦、余弦與正切公式1.Dcos2x=2cos2x1=2×3422.Dcos70°sin50°cos200°sin40°=cos70°sin50°+cos20°sin40°=cos70°sin50°+sin70°cos50°=sin(50°+70°)=sin120°=33.B因?yàn)棣痢师?3π2,且所以sinα=35,所以tanα=所以tanπ4.Asin2θ=cosπ=2sin2π4=2×1325.C因?yàn)閠anα=2tanπ5所以cos=sin=sin=sin=tanα+tanπ6.C∵cosα-π6+sinα=32cosα+32sin∴12cosα+32sin∴sinα+7π=32sin7.B∵0<y≤x<π2,且tanx=3tany,xy∈∴tan(xy)=tanx-tany當(dāng)且僅當(dāng)3tan2y=1時取等號,∴xy的最大值為π6,故選B8.0或12∵已知sin2α=22cos2α=22(12sin2α)=4sin2α∴2sinαcosα=4sin2α,∴sinα=0,或cosα=2sinα,即tanα=0,或tanα=19.-5π12,π12f(x)=sin2xsinπ6cos2xcos5π6=sin2xsinπ當(dāng)2kππ≤2xπ6≤2kπ(k∈即kπ5π12≤x≤kπ+π12(k∈Z)時,函數(shù)f(取k=0,得5π12≤故函數(shù)f(x)在-π210.133由C=60°,則A+B=120°,即A2+B2=60°.根據(jù)tanA2+B2=tanA解得tanA2tanB2=11.解(1)∵α,β∈∴π2<αβ<又tan(αβ)=13<∴π2<αβ<0由tan解得sin(αβ)=10(2)由(1)可得,cos(αβ)=3∵α為銳角,且sinα=35∴cosα=4∴cosβ=cos[α(αβ)]=cosαcos(αβ)+sinαsin(αβ)=412.D設(shè)△ABC的角A,B,C所對應(yīng)的邊分別為a,b,c.∵AB·∴bccosA=12bcsinA∴tanA=2,∴tan2A=2tanA1-tan213.D∵α∴sinα>0,cosα<0.∵3cos2α=sinπ4∴3(cos2αsin2α)=22(cosαsinα),∴cosα+sinα=2∴兩邊平方,可得1+2sinαcosα=118,∴sin2α=2sinαcosα=14.B∵3sin2θ=4tanθ,∴6sinθcosθsi∵θ≠kπ(k∈Z),tanθ≠0,∴31+tan2θ=2,解得tan∴cos2θ=cos2θsin2θ=cos2θ15.2令f(x)=4·1+cosx2·sinx2sinx|ln(x+1)|=sin2x|ln(x+1)|=0,即sin2x=|ln(x+1)|,在同一平面直角坐標(biāo)系中作出y=sin2x與y=|ln(x+由圖象知共有2個交點(diǎn),故f(x)的零點(diǎn)個數(shù)為2.16.Da=sin40°cos127°+cos40°sin127°=sin(40°+127°)=sin167°=sin13°,b=22(sin56°cos56°)=22sin56°22cos56°=sin(56°45°)=sinc=1=