(江苏专用)2020版高考数学大一轮复习第四章6第六节简单的三角恒等变换精练.docx

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1、1第六节 简单的三角恒等变换课时作业练1.(2018 江苏宝应中学第一次检测)已知 sin = , ,则 cos = . 35 (- 2, 2) ( +54 )答案 -210解析 sin = , ,35 (- 2, 2)cos = = ,1-sin245则 cos =cos =-cos( +54 ) +( + 4) ( + 4)=-cos cos +sin sin =- + =- . 4 4 45 22 35 22 2102.已知角 的顶点与原点 O 重合,始边与 x 轴的非负半轴重合,若它的终边与单位圆交于点 ,则 tan(45,35)= . (2 - 2)答案 -724解析 由三角函数定义

2、可得 tan = ,则 tan = = =- = =- .34 (2 - 2)sin(2 - 2)cos(2 - 2) -cos2sin2 1tan2 tan2 -12tan 7243.若锐角 , 满足 sin = ,tan(-)= ,则 tan = . 45 23答案 617解析 因为锐角 满足 sin = ,所以 cos = ,则 tan = = ,又 tan(-)= ,所以 tan =tan-45 35 sincos 43 23(-)= = .tan -tan( - )1+tan tan( - )6174.sin 10+cos 100,2x-1 1, 124.(2018 江苏海安高级中学高三月考)已知函数 y=cos x 与 y=sin(2x+)(00,得 x1 或 x-2 时, f(-2) f(t),即 mn.13e21

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