第一章1.3.2知能演练轻松闯关.doc

°+cos175°的值为( )° ° °解析:=cos5°-cos5°=(θ-)的值相等的式子为( )(+θ) (+θ)(π-θ) (π+θ)解析:(θ-)=-cosθ,经验证知只有sin(π+θ)=-(π+α)+cos(+α)=-m,则cos(-α)+2sin(6π-α)=( )A.- . D.-解析:选D.∵sin(π+α)+cos(+α)=-sinα-sinα=-m,∴sinα=.而cos(-α)+2sin(6π-α)=-sinα-2sinα=-3sinα=-.(x)=sinx,下列式子中成立的是( )(x+π)=sinx (2π-x)=(x-)=-cosx (π-x)=-f(x)解析:(x+π)=sin(x+π)=-sinx,f(2π-x)=sin(2π-x)=-sinx,f(x-)=sin(x-)=-sin(-x)=-cosx,f(π-x)=sin(π-x)=sinx=f(x),,2tan(π-α)-3cos(+β)+5=0,tan(π+α)+6sin(π+β)-1=0,则sinα的值是( )A. . :-2tanα+3sinβ+5=0,tanα-6sinβ-1=0.∴tanα=3,又tanα=,∴9==,sin2α=,∵α为锐角,∴sinα=,(+φ)=,且|φ|<,则tanφ=:cos(+φ)=-sinφ=.sinφ=-,又|φ|<,∴cosφ=,故tanφ=-.答案:-∈{4,5,6,7},且sin(-α)=-sinα,cos(-α)=cosα,:必须是偶数,∴k=:,化简=:原式==.∵α为第二象限角,∴sinα>0,cosα<0,∴原式==-:-:(1)·sin(α-)cos(+α);(2)sin(-α-5π)cos(α-)-sin(+α)cos(α-2π).解:(1)原式=·sin[-(-α)](-sinα)=·[-sin(-α)](-sinα)=·(-cosα)(-sinα)=-cos2α.(2)原式=sin(-α-π)cos[-(-α)]+cosαcos[-(2π-α)]=sin[-(α+π)]cos(-α)+cosαcos(2π-α)=-sin(α+π)sinα+cosαcosα=sin2α+cos2α=(x+)=-,求sin2(-x)-sin(-x):sin2(-x)-sin(-x)=sin2[-(x+)]-sin[π-(x+)]=cos2(x+)-sin(x+)=1-sin2(x+)-sin(x+