1、2两角和与差的三角函数2.1两角差的余弦函数2.2两角和与差的正弦、余弦函数课时过关能力提升1.在ABC中,sin Asin Bcos Acos B,则其形状是()A.锐角三角形B.钝角三角形C.直角三角形D.等腰三角形解析:sin Asin B0,即cos(A+B)0,cos C0.故这个三角形为钝角三角形.答案:B2.在ABC中,已知cos A=35,cos B=1517,则cos C等于()A.-1385B.1385C.-7785D.7785解析:因为cos A=35,cos B=1517,所以sin A=45,sin B=817,所以cos C=-cos(A+B)=-351517+45
2、817=-1385.答案:A3.函数f(x)=sin x-3cos x(x-,0)的递增区间是()A.-,-56B.-56,-6C.-3,0D.-6,0解析:f(x)=212sinx-32cosx=2sinxcos3-cosxsin3=2sinx-3.-x0,-43x-3-3,当-2x-3-3,即x-6,0时,f(x)是增加的.答案:D4.已知,均为锐角,且cos =1010,cos =55,则+的值是()A.23B.34C.4D.3解析:,均为锐角,sin =31010,sin =255.cos(+)=cos cos -sin sin =101055-31010255=-22.又,均为锐角,
3、0+.+=34.答案:B5.已知8sin +5cos =6,sin(+)=4780,则8cos +5sin =()A.10B.10C.-10D.20解析:设8cos +5sin =x,则(8sin +5cos )2+(8cos +5sin )2=62+x2,从而有64+25+80(sin cos +cos sin )=36+x2,89+804780=36+x2,x2=100,x=10.答案:A6.已知函数f(x)=3sin 2x+cos 2x,则下面结论中错误的是()A.函数f(x)的最小正周期为B.函数f(x)的图像可由g(x)=2sin 2x的图像向左平移6个单位长度得到C.函数f(x)的
4、图像关于直线x=6对称D.函数f(x)在区间0,6上是增加的解析:函数f(x)=3sin 2x+cos 2x=232sin2x+12cos2x=2sin2x+6,把g(x)=2sin 2x的图像向左平移6个单位长度得到函数y=2sin2x+6=2sin2x+3的图像,B选项是错误的.答案:B7.在ABC中,A=120,则sin B+sin C的最大值为.解析:由A=120,A+B+C=180,得sin B+sin C=sin B+sin(60-B)=32cos B+12sin B=sin(60+B).显然,当B=30时,sin B+sin C取得最大值1.答案:18.已知,34,sin(+)=
5、-35,sin-4=1213,则cos+4=.解析:由条件,得32+2,2-434,cos(+)=45,cos-4=-513.cos+4=cos(+)-4=cos(+)cos-4+sin(+)sin-4=45-513+-351213=-5665.答案:-56659.函数f(x)=sin(x+2)-2sin cos(x+)的最大值为.解析:f(x)=sin(x+2)-2sin cos(x+)=sin(x+)+-2sin cos(x+)=sin(x+)cos +cos(x+)sin -2sin cos(x+)=sin(x+)cos -cos(x+)sin =sin(x+)-=sin x.f(x)m
6、ax=1.答案:110.求证:sin(2+)sin-2cos(+)=sinsin.证明左边=sin+(+)-2sincos(+)sin=sincos(+)+cossin(+)-2sincos(+)sin=-sincos(+)+cossin(+)sin=sin(+)-sin=sinsin=右边,原等式成立.11.已知sin+3+sin =-435,-20,求cos 的值.解由已知,得sin cos3+cos sin3+sin =-435,32sin +32cos =-435,32sin +12cos =-45,即sin+6=-45.-20,-3+66,cos+6=35,cos =cos+6-6=
7、cos+6cos6+sin+6sin6=3532+-4512=33-410.12.已知a=(3,-1),b=(sin x,cos x),xR,f(x)=ab.(1)求f(x)的解析式;(2)求f(x)的周期、值域及单调区间.解(1)f(x)=ab=(3,-1)(sin x,cos x)=3sin x-cos x=232sinx-12cosx=2sinx-6(xR).(2)f(x)=2sinx-6,f(x)的周期为21=2,值域为-2,2.由-2+2kx-62+2k(kZ),得f(x)的递增区间为-3+2k,23+2k(kZ);由2+2kx-632+2k(kZ),得f(x)的递减区间为23+2k,53+2k(kZ).
