1、3.1.2两角和与差的正弦课时过关能力提升1.cos 23sin 53-sin 23cos 53等于()A.12B.-32C.-12D.32解析:原式=sin 53cos 23-cos 53sin 23=sin(53-23)=sin 30=12.答案:A2.如果2,且sin =45,那么sin+4-22cos 等于()A.225B.-225C.425D.-425解析:sin+4-22cos =sin cos4+cos sin4-22cos =22sin =2245=225.答案:A3.函数f(x)=5sin x-12cos x(xR)的最小值是()A.-5B.-12C.-13D.0解析:由于f
2、(x)=5sin x-12cos x=52+122sin(x+)=13sin(x+),其中,sin =-1213,cos =513.由于xR,所以x+R,故f(x)的最小值是-13.答案:C4.设a=2sin 24,b=sin 85-3cos 85,c=2(sin 47sin 66-sin 24sin 43),则()A.abcB.bcaC.cbaD.bac解析:b=sin 85-3cos 85=2sin(85-60)=2sin 25,c=2(sin 47sin 66-sin 24sin 43)=2(sin 47cos 24-cos 47sin 24)=2sin(47-24)=2sin 23,而
3、a=2sin 24,且sin 23sin 24ac.答案:D5.在ABC中,若sin B=2sin Acos C,则ABC一定是()A.等腰直角三角形B.等腰三角形C.直角三角形D.等边三角形解析:由于A+B+C=,所以B=-(A+C).于是sin B=sin-(A+C)=sin(A+C)=sin Acos C+cos Asin C,因此sin Acos C+cos Asin C=2sin Acos C,于是sin Acos C-cos Asin C=0,即sin(A-C)=0,必有A=C,ABC是等腰三角形.答案:B6.已知向量a=(cos x,sin x),b=(2,2),ab=85,则c
4、osx-4等于()A.-35B.-45C.35D.45解析:由ab=85,得2cos x+2sin x=85,22cos x+22sin x=45,即cos x-4=45,故选D.答案:D7.若,都为锐角,则sin(+)与sin +sin 的值满足()A.sin(+) sin +sin B.sin(+)sin +sin C.sin(+)=sin +sin D.sin(+)sin +sin 解析:将sin(+)利用两角和的正弦公式展开,注意锐角条件,则有sin(+)=sin cos +cos sin sin +sin .答案:B8.已知tan(+)=2,则sincos+cossin+cos(+)
5、sin(+)-coscos+sinsin=.解析:原式=sin(+)+cos(+)sin(+)-cos(+)=tan(+)+1tan(+)-1=2+12-1=3.答案:39.要使sin -3cos =2m+1有意义,则m的取值范围是.解析:由于sin -3cos =212sin-32cos=2sin-3,因此-22m+12,即-32m12.答案:-32,1210.已知cos4-=35,sin34+=513,其中434,04,求sin(+)的值.解:+2=34+-4-,sin(+)=-cos2+(+)=-cos 34+-4-=-cos34+cos4-sin34+sin4-.434,04,-24-0,3434+.sin4-=-45,cos34+=-1213.sin(+)=-121335-513-45=5665.11.已知函数f(x)=-1+2sin 2x+mcos 2x的图象经过点A(0,1),求此函数在0,2上的最值.解:点A(0,1)在函数f(x)的图象上,1=-1+2sin 0+mcos 0,解得m=2.f(x)=-1+2sin 2x+2cos 2x=2(sin 2x+cos 2x)-1=22sin2x+4-1.0x2,42x+454.-22sin2x+41.-3f(x)22-1.函数f(x)的最大值为22-1,最小值为-3.
