3、∈Z时,sin(kπ-α)cos(kπ+α)sin[(k+1)π+α]cos[(k+1)π-α]=( )A.-1B.1C.-1或1D.0答案 A 若k为偶数,则原式=sin(-α)cosαsin(π+α)cos(π-α)=-sinαcosα(-sinα)(-cosα)=-1;若k为奇数,则原式=sin(π-α)cos(π+α)sinαcos(-α)=sinα(-cosα)sinαcosα=-1.故选A. 5.(2016课标全国Ⅲ文,6,5分)若tanθ=-13,则cos2θ=( )A.-45B
4、.-15C.15D.45答案 D 解法一:cos2θ=cos2θ-sin2θ=cos2θ-sin2θcos2θ+sin2θ=1-tan2θ1+tan2θ,∵tanθ=-13,∴cos2θ=45.故选D.解法二:由tanθ=-13,可得sinθ=±110,因而cos2θ=1-2sin2θ=45.6.(2019镇海中学月考)若θ∈π2,π,且2cos2θ=sinπ4-θ,则sin2θ=( )A.14B.-14C.34D.-34答案 D 由2cos2θ=sinπ4-θ,得2(cos2θ-sin2θ)=22·(cosθ-sinθ),又θ
5、∈π2,π,则cosθ-sinθ≠0,得cosθ+sinθ=12,两边平方得cos2θ+sin2θ+2sinθcosθ=14,即sin2θ=-34.7.已知θ为钝角,且sinθ+cosθ=15,则tan2θ=( )A.-247B.247C.-724D.724答案 B 由sinθ+cosθ=15得(sinθ+cosθ)2=125,即2sinθcosθ=-2425,亦即sin2θ=-2425.因为θ为钝角,所以θ∈π2,π,所以2θ∈(π,2π),cos2θ=-725,所以tan2θ=247,故选B.8.(2019效实中学月考)已知
6、2sin(π+α)-cosπ2-αsin(-α)-cos(π+α)=4,则tanα= . 答案 4解析 2sin(π+α)-cosπ2-αsin(-α)-cos(π+α)=-2sinα-sinα-sinα+cosα=-3sinαcosα-sinα=4,即4cosα-4sinα=-3sinα,∴4cosα=sinα,∴tanα=sinαcosα=4.9.已知sinα+2cosα=0,则2sinαcosα-cos2α的值是 . 答案 -1解析 由sinα+2cosα=0得tanα=-2.2sinαcosα-cos2α=2s
7、inαcosα-cos2αsin2α+cos2α=2tanα-1tan2α+1=2×(-2)-1(-2)2+1=-55=-1.10.若α∈0,π2,且sin2α+cos2α=14,则cosα= ,tanα= . 答案 12;3解析 由sin2α+cos2α=14,得sin2α+1-2sin2α=1-sin2α=cos2α=14,因为α∈0,π2,所以cosα=12,所以α=π3,故tanα=3.11.1-sin6x-cos6x1-sin4x-cos4x= . 答案 32解析 sin6x+cos6x=(sin2x+
8、cos2x)(sin4x-sin2x·cos2x+cos4x)=(sin2x+cos2x)2-3sin2x·cos2x=1-3sin2x·cos2x.sin4x+cos4x=(sin2x+cos2x)2-2sin2x·cos2x=1-2sin2x·
