全国2018-2019高中数学第三章三角恒等变换3.1和角公式3.1.2两角和与差的正弦练习新人教B版.docx

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.B.-C.-D.解析:原式=sin 53cos 23-cos 53sin 23=sin(53-23)=sin 30=.答案:A2.如果,且sin =,那么sincos 等于()A.B.-C.D.-解析:sincos =sin cos+cos sincos =sin =.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(x)=5sin x-12cos x=sin(x+)=13sin(x+),其中,sin =-,cos =.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-cos 85,c=2(sin 47sin 66-sin 24sin 43),则()A.abcB.bcaC.cbaD.bac解析:b=sin 85-cos 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=(),ab=,则cos等于()A.-B.-C.D.解析:由ab=,得cos x+sin x=,cos x+sin x=,即cos ,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=.解析:原式=3.答案:39.要使sin -cos =2m+1有意义,则m的取值范围是.解析:由于sin -cos =2=2sin,因此-22m+12,即-m.答案:10.已知cos,sin,其中,0,求sin(+)的值.解:+-,sin(+)=-cos=-cos =-coscos-sinsin.,0,-0,+.sin=-,cos=-.sin(+)=-.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