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课时规范练21三角恒等变换基础巩固组1.函数f(x)=(3sin x+cos x)(3cos x-sin x)的最小正周期是()A.2B.C.32D.22.已知sin+5=33,则cos2+25=()A.B.33C.D.323.(2018云南民族中学一模)已知tan =2,则2sin2+1cos2-4的值是()A.B.-134C.135D.1344.(2018四川成都七中模拟)已知sin76+=33,则cos23-2=()A.-B.-C.D.5.已知f(x)=sin2x+sin xcos x,则f(x)的最小正周期和一个递增区间分别为()A.,0,B.2,-4,34C.,-8,38D.2,-4,46.(2018黑龙江高考仿真(三)已知sin+3+sin =-435,则cos+23=()A.-B.-C.D.7.(2018全国第一次大联考)已知sinx+3=13,则sin53-x-cos2x-3的值为.8.设f(x)=1+cos2x2sin2-x+sin x+a2sinx+4的最大值为2+3,则实数a=.9.设为锐角,若cos+6=45,则sin2+12的值为.10.(2018湖北百所重点校联考)设0,3,满足6sin +2cos =3.(1)求cos+6的值;(2)求cos2+12的值.综合提升组11.已知函数f(x)=sin(x+)+10,02的图像的相邻两对称轴之间的距离为,且在x=6时取得最大值2,若f()=,且623,则sin2+23的值为()A.1225B.-1225C.2425D.-242512.已知-3,0,cos+6-sin =435,则sin+12的值是()A.-235B.-210C.235D.-13.(2018湖南长郡中学一模,17改编)已知函数f(x)=2sin xcos22+cos xsin -sin x(0)在x=处取最小值.则的值为.14.(2018安徽合肥二模)已知a=(sin x,3cos x),b=(cos x,-cos x),函数f(x)=ab+32.(1)求函数y=f(x)图像的对称轴方程;(2)若方程f(x)=在(0,)上的解为x1,x2,求cos(x1-x2)的值.创新应用组15.已知m=tan(+)tan(-+),若sin 2(+)=3sin 2,则m=()A.-1B.C.D.216.函数y=sin +cos -4sin cos +1,且2sin2+sin21+tan=k,42,(1)把y表示成k的函数f(k);(2)求f(k)的最大值.课时规范练21三角恒等变换1.Bf(x)=2sinx+62cosx+6=2sin2x+3,故最小正周期T=22=,故选B.2.A由题意sin+5=33,cos2+25=cos 2+5=1-2sin2+5=1-2332=13.故选A.3.Dtan =2,2sin2+1cos2-4=2sin2+sin2+cos2cos2-2=3sin2+cos2sin2=3sin2+cos22sincos=3tan2+12tan=322+122=134.4.B由题意sin76+=sin+6+=-sin6+,所以sin6+=-33,由于cos23-2=cos-3+2=-cos3+2=-cos26+=2sin26+-1=2-332-1=-13,故选B.5.C由f(x)=sin2x+sin xcos x=1-cos2x2+12sin 2x=12+2222sin2x-22cos2x=12+22sin2x-4,则T=22=.又2k-22x-42k+2(kZ),k-8xk+38(kZ)为函数的递增区间.故选C.6.Dsin3+sin =sincos +cossin +sin =-435,32sin +32cos =-435,即32sin +12cos =-45.sin+6=-45.故cos+23=cos+2+6=-sin+6=45.7.sin53-x-cos2x-3=sin2-x+3-cos 2x+3-2=-sinx+3+cos 2x+3=-sinx+3+1-2sin2x+3=-+1-29=49.8.3f(x)=1+2cos2x-12cosx+sin x+a2sinx+4=cos x+sin x+a2sinx+4=2sinx+4+a2sinx+4=(2+a2)sinx+4.依题意有2+a2=2+3,则a=3.9.17502为锐角,cos+6=45,sin+6=35,sin2+3=2sin+6cos+6=2425,cos2+3=2cos2+6-1=725,sin2+12=sin2+3-4=22sin2+3-cos2+3=17250.10.解 (1)6sin +2cos =3,sin+6=64.0,3,+66,2,cos+6=104.(2)由(1)可得cos2+3=2cos2+6-1=21042-1=14.0,3,2+33,sin2+3=154.cos2+12=cos2+3-4=cos2+3cos4+sin2+3sin4=30+28.11.D由题意,T=2,即T=2=2,即=1.又当x=6时,f(x)取得最大值,即6+=2+2k,kZ,即=3+2k,kZ.02,=3,f(x)=sinx+3+1.f()=sin+3+1=95,可得sin+3=45.623,可得2+3,cos+3=-35.sin2+23=2sin+3cos+3=245-35=-2425.故选D.12.B由cos+6-sin =435,可得32cos -32sin =435,12cos -32sin =45,cos+3=45.-3,0,+30,3,sin+3=35,sin+12=sin+3-4=22sin+3-22cos+3=2235-45=-210,故选B.13.f(x)=2sin x1+cos2+cos xsin -sin x=sin x+sin xcos +cos xsin -sin x=sin xcos +cos xsin =sin(x+).因为函数f(x)在x=处取最小值,所以sin(+)=-1,由诱导公式知sin =1,因为00,且0x1512x223,易知(x1,f(x1)与(x2,f(x2)关于x=512对称,则x1+x2=56,cos(x1-x2)=cosx1-56-x1=cos2x1-56=cos2x1-3-2=sin2x1-3=13.15.Dsin 2(+)=3sin 2,sin(+)-(-)=3sin(+)-(+-),sin(+)cos(-)-cos(+)sin(-)=3sin(+)cos(+-)-3cos(+)sin(+-),即-2sin(+)cos(+-)=-4cos(+)sin(+-),12tan(+)=tan(+-),故m=tan(+)tan(-+)=2,故选D.16.解 (1)k=2sin2+sin21+tan=2sin2+2sincos1+sincos=2sin(sin+cos)cos+sincos=2sin cos ,(sin +cos )2=1+2sin cos =1+k.40.sin +cos =1+k.y=1+k-2k+1.由于k=2sin cos =sin 2,42,0k1.f(k)=1+k-2k+1(0k1).(2)设1+k=t,则k=t2-1,1t2.y=t-(2t2-2)+1,即y=-2t2+t+3(1t2).关于t的二次函数在区间1,2)内是减少的,t=1时,y取最大值2.
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