2020届高考数学一轮复习第4章三角函数、解三角形16同角三角函数关系式及诱导公式课时训练文(含解析).doc

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2020届高考数学一轮复习第4章三角函数、解三角形16同角三角函数关系式及诱导公式课时训练文(含解析).doc_第1页
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1、1【课时训练】同角三角函数关系式及诱导公式一、选择题1(2018 大庆一中期末)已知 ,sin ,则 cos( )的值为( )( 2, 0) 35A B45 45C D35 35【答案】A【解析】 ,sin ,cos .( 2, 0) 35 45cos( )cos .故选 A.452(2019 广东江门调研)sin ( )113A B32 32C D12 12【答案】B【解析】sin sin sin sin ,故选 B.113 (4 3) ( 3) 3 323(2018 长春第一次调研) ( )1 2sin 2 cos 2Asin 2cos 2 Bsin 2cos 2C(sin 2cos 2)

2、 Dcos 2sin 2【答案】A【解析】 1 2sin 2 cos 2 1 2sin 2cos 2 sin 2 cos 2 2|sin 2cos 2|sin 2cos 2,故选 A.4(2018 重庆凤鸣山中学月考)若 为三角形的一个内角,且 sin cos ,23则这个三角形是( )A正三角形 B直角三角形C锐角三角形 D钝角三角形【答案】D【解析】由 sin cos ,得(sin cos )2 ,23 4912sin cos ,2sin cos .49 592 (0,), 为钝角故选 D.5(2018 陕西西安模拟)已知 cos ,则 sin ( )( 6 ) 23 ( 23)A B23

3、 12C D23 12【答案】A【解析】sin sin( 23) 2 ( 6 )sin cos . 2 ( 6 ) ( 6 ) 23故选 A.6(2018 东北三校联合模拟)已知 sin cos ,且 ,则 cos 18 54 32 sin 的值为( )A B32 32C D34 34【答案】B【解析】 ,cos 0,sin 0 且 cos sin .54 32cos sin 0.又(cos sin )212sin cos 12 ,18 34cos sin .327(2018 武汉模拟)已知 ,sin cos ,则 tan ( )( 2, ) 15 ( 4)A7 B7C D17 17【答案】C【解析】由 sin cos 两边平方,得 12sin cos ,2sin 15 125 cos .又 0,cos 0,cos A0,120169 289169sin Acos A .1713由,可知 sin A ,cos A ,1213 513tan A .sin Acos A1213 513 125

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