5.5.1两角和与差的正弦、余弦和正切公式第一课时同步练习—2022-2023学年高一上学期数学人教A版（2019）必修第一册.docx

1、两角和与差的正弦、余弦和正切公式第一课时 练习1.cos 20=().A.cos 30cos 10-sin 30sin 10B.cos 30cos 10+sin 30sin 10C.sin 30cos 10-sin 10cos 30D.cos 30cos 10-sin 30cos 102.若cos 5xcos(-2x)-sin(-5x)sin 2x=0,则x的值可能是().A.10 B.6 C .5 D.43.若cos(-)=13,则(sin +sin )2+(cos +cos )2=().A.83 B.-83 C.223 D.-2234.(多选题)已知,是锐角,cos =55,cos(-)=

2、31010,则cos =().A.22 B.7210 C.210 D.-225.在ABC中,C=90,AC=3,BC=4,则cos(A-B)的值是().A.35 B.45 C.2425 D.7256.已知,均为锐角,满足cos =255,sin =1010,则cos(-)=.7.函数f(x)=sin x+cos x的单调递减区间为,最大值为.8.(多选题)已知,0,2,sin +sin =sin ,cos +cos =cos ,则下列说法正确的是().A.cos(-)=12B.cos(-)=-12C.-=3D.-=-39.若sin sin =1,则cos(-)=().A.0 B.1 C.1 D

3、.-110.化简:2cos10-sin20cos20=.11.已知A(cos ,sin ),B(cos ,sin ),其中,为锐角,且|AB|=105.(1)求cos(-)的值;(2)若cos =35,求cos 的值.12.已知cos(-)=-1213,cos(+)=1213,且-2,+32,2,求角的值.参考答案1.B2.B3.A4.AC5.C6.72107.2k+4,2k+54,kZ28.AC9.B10.311.【解析】(1)由|AB|=105,得(cos-cos)2+(sin-sin)2=105,2-2(cos cos +sin sin )=25,cos(-)=45.(2)cos =35

4、,cos(-)=45,为锐角,sin =45,sin(-)=35.当sin(-)=35时,cos =cos-(-)=cos cos(-)+sin sin(-)=3545+4535=2425;当sin(-)=-35时,cos =cos-(-)=cos cos(-)+sin sin(-)=3545+45-35=0.为锐角,cos =2425. 12.【解析】由-2,且cos(-)=-1213,得sin(-)=513.由+32,2,且cos(+)=1213,得sin(+)=-513,cos 2=cos(+)-(-)=cos(+)cos(-)+sin(+)sin(-)=1213-1213+-513513=-1.又-2,+32,2,22,32,2=,则=2.