新教材高考数学一轮复习考点规范练21三角恒等变换含解析新人教版

PAGEPAGE6考点规范练21三角恒等变换一、基础巩固1.已知α∈π,3π2,且cosα=-45A.7 B.17 C.-17 D.2.2sin47°-3sin17A.-3 B.-1 C.3 D.13.(多选)下列选项中,与sin-11π6A.2cos215°-1B.cos18°cos42°-sin18°sin42°C.2sin15°sin75°D.tan304.已知cosα-π6+sinα=435A.12 B.32 C.-45 D5.已知2sin2α=1+cos2α,则tan2α等于()A.43 B.-43 C.43或0 D.6.已知5sin2α=6cosα,α∈0,π2,则tanαA.-23 B.13 C.357.若0<y≤x<π2,且tanx=3tany,则x-y的最大值为(A.π4 B.π6 C.π38.若sinα-π4=12,则cosπ2+2α等于()A.-34 B.-23 C.-12 D9.(2020浙江,13)已知tanθ=2,则cos2θ=;tanθ-π4=10.设函数f(x)=1+cos2x2sinπ2-x+sinx+a2sinx+π4的最大值为2+11.已知函数f(x)=cos-x2+sinπ-x2(1)求f(x)的最小正周期;(2)若f(α)=2105,α∈0,π二、综合应用12.设a=cos50°cos127°+cos40°cos37°,b=22(sin56°-cos56°),c=1-tan239°1+tanA.a>b>c B.b>a>c C.c>a>b D.a>c>b13.(多选)化简下列各式,与tanα相等的是()A.1B.1+cos(π+2α)2C.1D.sin214.已知sin10°+mcos10°=2cos140°,则m=.

15.已知函数f(x)=2sinx+5π24cosx+5π24-2cos2x+5π24+1,则f(x)的最小正周期为三、探究创新16.已知函数f(x)=cosωx(sinωx+3cosωx)(ω>0),若存在实数x0,使得对任意的实数x,都有f(x0)≤f(x)≤f(x0+2020π)成立,则ω的最小值为()A.12020π B.14032π C.17.已知函数f(x)=1+1tanxsin2x-2sinx+π4·(1)若tanα=2,求f(α)的值;(2)若x∈π12,π2,求f

考点规范练21三角恒等变换1.B因为α∈π,3π所以sinα=-35,所以tanα=所以tanπ2.D原式=2×sin47°-sin17°cos30°cos17°=2×3.BCsin-11π6A选项中,2cos215°-1=cos30°=32B选项中,cos18°cos42°-sin18°sin42°=cos(18°+42°)=cos60°=12C选项中,2sin15°sin75°=2sin15°cos15°=sin30°=12D选项中,tan30°+tan15°1-4.C∵cosα-π6+sinα=32cosα+32∴12cosα+32sin则sinα+7π6=-sinα+π6=-(32sin5.C因为2sin2α=1+cos2α,所以2sin2α=2cos2α.所以2cosα(2sinα-cosα)=0,解得cosα=0或tanα=1若cosα=0,则α=kπ+π2(k∈Z),2α=2kπ+π(k∈Z),所以tan2α=0;若tanα=12,则tan2α综上所述,故选C.6.B由题意,知10sinαcosα=6cosα,∵α∈0,π2,∴sinα=3∴tanα7.B∵0<y≤x<π2,∴x-y又tanx=3tany,∴tan(x-y)=tanx-tany1+tanxtany=2tany1+3tan8.C∵sin(α-π4)=1∴cosπ2+2α=-cos[π-π2+2α]=-cos(π2-2α)=-cos2α-π2=-1+2sin2(9.-3513cos2θ=cos2θ-sin2θ=co10.±3f(x)=1+2cos2x-12cos=cosx+sinx+a2sinx=2sinx+π4+a=(2+a2)sinx依题意有2+a2=2+3,则a=±311.解(1)f(x)=cos-x2+sinπ-x2=sinx2+cosx2=2sinx2+(2)由f(α)=2105,得sinα2+cosα2=2105,则sinα2+cosα22=21052,即1+sinα=812.Da=sin40°cos127°+cos40°sin127°=sin(40°+127°)=sin167°=sin13°,b=22(sin56°-cos56°)=22sin56°-22cos56°=sin(56°-c=1-tan239°1+tan239°=co∵sin13°>sin12°>sin11°,∴a>c>b.故选D.13.BC对于A,1-cos2α1+cos2α=1-(1-2sin2α)1+2cos2α-1=sin2αcos2α=ta对于B,因为α∈(0,π),所以1+cos(π+2α对于C,1-cos2αsin2对于D,sin2α1-cos214.-3由sin10°+mcos10°=2cos140°,可得m=2cos140°-sin1015.πkπ-π3,k∵f(x)=2sin(x+5π24)·cos(x+5π24)-2cos2x+5π24+1=sin=2=2sin2x+5∴f(x)的最小正周期T=2π2=由f(x)=2sin2x+π6,得当2kπ-π2≤2x+π6≤2kπ+π2(k∈Z),即kπ-π3≤x≤kπ+π6(k∈Z)时,f(x)单调递增,即函数f(x)的单调递增区间是[kπ-π16.C由题意可得,f(x0)是函数f(x)的最小值,f(x0+2020π)是函数f(x)的最大值.因为f(x)=cosωx(sinωx+3cosωx)=12sin2ωx+3×1+cos2ωx2=sin2ωx+π3+32,所以要使故2020π=12×2πωmin,求得ωmin=117.解(1)f(x)=sin2x+sinxcosx+2sinx+π4·cosx=12+12(sin2x-c