全国通用2019届高考数学大一轮复习第四章三角函数解三角形4.5简单的三角恒等变换第2课时课件.ppt
第2课时 简单的三角恒等变换,§4.5 简单的三角恒等变换,课时作业,题型分类 深度剖析,内容索引,题型分类 深度剖析,题型一 三角函数式的化简,自主演练,答案,解析,4sin ,解析,答案,解析,答案,解析,答案,为第二象限角,,(1)三角函数式的化简要遵循“三看”原则:一看角,二看名,三看式子结构与特征. (2)三角函数式的化简要注意观察条件中角之间的联系(和、差、倍、互余、互补等),寻找式子和三角函数公式之间的共同点.,典例 (1)(2018·太原质检)2sin 50°sin 10°(1 tan 10°)· .,题型二 三角函数的求值,多维探究,命题点1 给角求值与给值求值,答案,解析,答案,解析,答案,解析,cos cos() cos()cos sin()sin ,命题点2 给值求角,解析,答案,答案,解析,解析 tan tan(),答案,解析,cos()cos cos sin sin ,(1)给角求值与给值求值问题的关键在“变角”,通过角之间的联系寻找转化方法. (2)给值求角问题:先求角的某一三角函数值,再求角的范围确定角.,答案,解析,则(2sin 3cos )·(sin cos )0,,2sin 3cos ,又sin2cos21,,解析,4,答案,解析,答案,于是sin sin() sin cos()cos sin(),典例 (2017·浙江)已知函数f(x)sin2xcos2x2 sin xcos x(xR). (1)求 的值;,题型三 三角恒等变换的应用,师生共研,解答,解 由cos 2xcos2xsin2x与sin 2x2sin xcos x,得,所以f(x)的最小正周期是. 由正弦函数的性质,得,(2)求f(x)的最小正周期及单调递增区间.,解答,三角恒等变换的应用策略 (1)进行三角恒等变换要抓住:变角、变函数名称、变结构,尤其是角之间的关系;注意公式的逆用和变形使用. (2)把形如yasin xbcos x化为y sin(x),可进一步研究函数的周期性、单调性、最值与对称性.,答案,解析,跟踪训练 (1)函数f(x)sin(x)2sin cos x的最大值为 .,1,解析 因为f(x)sin(x)2sin cos x sin xcos cos xsin sin(x), 又1sin(x)1, 所以f(x)的最大值为1.,解析,答案,化归思想和整体代换思想在三角函数中的应用,思想方法,思想方法指导,思想方法指导 (1)讨论形如yasin xbcos x型函数的性质,一律化成y sin(x)型的函数. (2)研究yAsin(x)型函数的最值、单调性,可将x视为一个整体,换元后结合ysin x的图象解决.,规范解答,规范解答,课时作业,基础保分练,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,即sin cos cos cos sin ,,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析 f(x)5cos x12sin x,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,tan 0且tan 0,,即0,结合tan()1,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析 cos4sin4(sin2cos2)(cos2sin2),2(0,),,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,12.已知函数f(x)cos2xsin xcos x,xR. (1)求 的值;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,技能提升练,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,14.在斜ABC中,sin A cos Bcos C,且tan B·tan C1 , 则角A的值为 .,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,拓展冲刺练,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,16.(2018·泉州模拟)已知角的顶点在坐标原点,始边与x轴的正半轴重合,终边经过点P(3, ). (1)求sin 2tan 的值;,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,sin 2tan 2sin cos tan ,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解 f(x)cos(x)cos sin(x)sin cos x,xR,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,