1、3.2.2半角的正弦、余弦和正切课时过关能力提升1.若sin =,则sin的值等于()A.B.-C.D.-解析:由sin =可得cos =-.又,所以sin.答案:C2.tan 15+cot 15等于()A.2B.2C.4D.解析:tan 15+cot 15=4.答案:C3.设(,2),则等于()A.sinB.cosC.-sinD.-cos解析:由(,2)知,所以=sin.答案:A4.若,则sin +cos 的值是()A.B.C.1D.解析:由,结合sin2+cos2=1可得sin =(sin =0舍去),于是cos =,从而sin +cos =.答案:A5.若,sin 2=,则sin 等于(
2、)A.B.C.D.解析:由,得2.又sin 2=,故cos 2=-.故sin =.答案:D6.化简等于()A.tan 2B.cot 4C.tan 4D.cot 2解析:=tan 4.答案:C7.已知为三角形的内角,sin =,则tan=.解析:由已知得cos =,且,于是tan=3或.答案:3或8.若2,且cos =,则的值是.解析:.答案:9.已知090,sin 与sin 是方程x2-(cos 40)x+cos240-=0的两根,则cos(2-)=.解析:由已知,得=2cos240-4cos240+2=2sin240,x=cos 40sin 40.x1=sin 45cos 40+cos 45
3、sin 40=sin 85,x2=sin 45cos 40-cos 45sin 40=sin 5.又由090,知=85,=5,cos(2-)=cos(-75)=cos 75=cos(45+30)=.答案:10.已知sinsin,求2sin2+tan -1的值.解:sinsin,2sincos,即sin.cos 4=.而2sin2+tan -1=-cos 2+=-.,2.cos 2=-=-,tan 2=-=-.-=-,即2sin2+tan -1的值为.11.已知向量a=(sin x,-cos x),b=(cos x,cos x),函数f(x)=ab+.(1)求f(x)的最小正周期;(2)当0x时,求函数f(x)的值域.解:(1)f(x)=sin xcos x-cos2x+=sin 2x-(cos 2x+1)+=sin 2x-cos 2x=sin.故f(x)的最小正周期为.(2)0x,-2x-,-sin1,即f(x)的值域为.
