全国通用版2018-2019高中数学第三章三角恒等变换3.2.2半角的正弦余弦和正切练习新人教B版必修4 .doc

3.2.2半角的正弦、余弦和正切课时过关能力提升1.若sin =513,2,则sin的值等于()A.2626B.-2626C.52626D.-52626解析:由sin =513且2可得cos =-1213.又422,所以sin2=1-cos2=1-12132=2526=52626.答案:C2.tan 15+cot 15等于()A.2B.23C.4D.433解析:tan 15+cot 15=1-cos30sin30+sin301-cos30=4.答案:C3.设(,2),则1+cos(+)2等于()A.sin2B.cos2C.-sin2D.-cos2解析:由(,2)知22,所以1+cos+2=1-cos2=sin2=sin2.答案:A4.若sin1+cos=12,则sin +cos 的值是()A.B.C.1D.2915解析:由sin1+cos=12,结合sin2+cos2=1可得sin = (sin =0舍去),于是cos =,从而sin +cos =.答案:A5.若4,2,sin 2=378,则sin 等于()A.B.C.74D.解析:由4,2,得22,.又sin 2=378,故cos 2=-18.故sin =1-cos22=34.答案:D6.化简1+sin8-cos81+sin8+cos8等于()A.tan 2B.cot 4C.tan 4D.cot 2解析:1+sin8-cos81+sin8+cos8=(1-cos8)+sin8(1+cos8)+sin8=2sin24+2sin4cos42cos24+2sin4cos4=2sin4(sin4+cos4)2cos4(cos4+sin4)=sin4cos4=tan 4.答案:C7.已知为三角形的内角,sin =,则tan2=.解析:由已知得cos =,且20,2,于是tan2=1-cos1+cos=3或.答案:3或8.若322,且cos =,则12+1212+12cos2的值是.解析:12+1212+12cos2=12+12cos2=12+12cos=12+1214=104.答案:1049.已知090,sin 与sin 是方程x2-(2cos 40)x+cos240-=0的两根,则cos(2-)=.解析:由已知,得=2cos240-4cos240+2=2sin240,x=22cos 4022sin 40.x1=sin 45cos 40+cos 45sin 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)=6-24.答案:6-2410.已知sin4+2sin4-2=14,4,2,求2sin2+tan -1tan-1的值.解:sin4+2sin4-2=14,2sin4+2cos4+2=12,即sin2+4=12.cos 4=12.而2sin2+tan -1tan-1=-cos 2+sin2-cos2sincos=-cos2+2tan2.4,2,22,.cos 2=-1+cos42=-32,tan 2=-1-cos41+cos4=-33.-cos2+2tan2=-32+2-33=523,即2sin2+tan -1tan-1的值为532.11.已知向量a=(sin x,-cos x),b=(cos x,3cos x),函数f(x)=ab+32.(1)求f(x)的最小正周期;(2)当0x2时,求函数f(x)的值域.解:(1)f(x)=sin xcos x-3cos2x+32=12sin 2x-32(cos 2x+1)+32=12