1、3.2简单的三角恒等变换1.cos2的值为()A.B.C.D.解析:cos2.答案:B2.已知cos,540720,则sin等于()A.B.C.-D.-解析:540720,270360,135180.sin.答案:A3.已知2sin =1+cos ,则tan等于()A.B.或不存在C.2D.2或不存在解析:由2sin =1+cos ,得4sincos=2cos2.当cos=0时,tan不存在,当cos0时,tan.答案:B4.已知tan=3,则cos =()A.B.-C.-D.解析:cos =cos2-sin2=-.答案:B5.若函数f(x)=(1+tan x)cos x,则f=()A.B.-
2、C.1D.解析:f(x)=cos x=cos x+sin x=2sin,f=2sin=2sin.答案:D6.若cos =-,是第三象限的角,则等于()A.-B.C.2D.-2解析:是第三象限角,cos =-,sin =-.=-.答案:A7.设56,cos =a,则sin =.解析:56,3,.又cos =a,sin =-=-=-.答案:-8.在ABC中,若cos A=,则sin2+cos 2A等于.解析:在ABC中,B+C=-A,sin2+cos 2A=+cos 2A=+cos 2A=+2cos2A-1=2cos2A+cos A-=-.答案:-9.导学号08720093已知sin,0,则的值等
3、于.解析:sin,0,cos.=.答案:10.已知函数f(x)=sin (1-a)x+cos (1-a)x的最大值为2,则f(x)的最小正周期为.解析:f(x)=sin (1-a)x+,由已知得=2,a=3.f(x)=2sin(-2x+).T=.答案:11.求-sin 10的值.解:原式=-sin 10=.12.导学号08720094已知2sin=sin +cos ,2sin2=sin 2,求证:sin 2+cos 2=0.证明:2sin=sin +cos ,(sin +cos )=sin +cos .两边平方得2(1+sin 2)=1+sin 2,sin 2=1+2sin 2.又sin 2=2sin2,sin 2=1-cos 2.1-cos 2=1+2sin 2.2sin 2+cos 2=0,sin 2+cos 2=0.
