1、5.4 简单的三角恒等变换,第五章 三角函数、解三角形,NEIRONGSUOYIN,内容索引,基础知识 自主学习,题型分类 深度剖析,课时作业,1,基础知识 自主学习,PART ONE,知识梳理,1.两角和与差的余弦、正弦、正切公式,ZHISHISHULI,cos cos sin sin ,sin cos cos sin ,sin cos cos sin ,cos()cos cos sin sin (C() cos() (C() sin() (S() sin() (S(),2.二倍角公式,sin 2 ； cos 2 ；tan 2 .,2sin cos ,12sin2,cos2sin2,2cos
2、21,1.诱导公式与两角和差的三角函数公式有何关系？,【概念方法微思考】,提示 诱导公式可以看成和差公式中k (kZ)时的特殊情形.,2.怎样研究形如f(x)asin xbcos x函数的性质？,提示 先根据辅助角公式asin xbcos x sin(x)，将f(x)化成f(x)Asin(x)k的形式，再结合图象研究函数的性质.,基础自测,JICHUZICE,题组一 思考辨析 1.判断下列结论是否正确(请在括号中打“”或“”) (1)存在实数，使等式sin()sin sin 成立.( ),1,2,3,4,5,6,7,8,题组二 教材改编,1,2,3,4,5,6,7,8,1,2,3,4,5,6,
3、7,3.P131T5sin 347cos 148sin 77cos 58 .,解析 sin 347cos 148sin 77cos 58 sin(27077)cos(9058)sin 77cos 58 (cos 77)(sin 58)sin 77cos 58 sin 58cos 77cos 58sin 77,8,4.P146A组T4(2)tan 10tan 50 tan 10tan 50 .,1,2,3,4,5,6,7,8,题组三 易错自纠,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7
4、,8,1,2,3,4,5,6,7,4sin ,8,2,题型分类 深度剖析,PART TWO,第1课时 两角和与差的正弦、余弦和正切公式,题型一 和差公式的直接应用,自主演练,解析 sin215cos215(cos215sin215),(1)使用两角和与差的三角函数公式，首先要记住公式的结构特征. (2)使用公式求值，应先求出相关角的函数值，再代入公式求值.,题型二 和差公式的灵活应用,多维探究,命题点1 角的变换,于是cos cos() cos()cos sin()sin ,命题点2 三角函数式的变换,命题点3 公式的逆用与变形,得sin22sin cos cos2sin22sin cos c
5、os2 (sin2cos2)(cos2sin2)2(sin cos sin cos )112sin(),(1)解决三角函数的求值问题的关键是把“所求角”用“已知角”表示.当“已知角”有两个时，“所求角”一般表示为两个“已知角”的和或差的形式；当“已知角”有一个时，此时应着眼于“所求角”与“已知角”的和或差的关系.,sin sin()sin()cos cos()sin ,思想方法,SIXIANGFANGFA,用联系的观点进行三角变换,三角变换的关键是找到条件和结论中的角和式子结构之间的联系.变换中可以通过适当地拆角、凑角或对式子整体变形达到目的.,例 (1)(2018绍兴一中期中)(1tan 2
6、1)(1tan 20)(1tan 25)(1tan 24)的值为 A.2 B.4 C.8 D.16,解析 (1tan 21)(1tan 20)(1tan 25)(1tan 24) 1tan(4524)(1tan 24)1tan(4525)(1tan 25),cos sin ，,3,课时作业,PART THREE,基础保分练,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,A.abc B.bac C.cab D.acb,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
7、,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析 asin 40cos 127cos 40sin 127 sin(40127)sin 167sin 13，,sin(5645)sin 11，,sin 13sin 12sin 11，acb.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,
8、3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以cos()cos2() cos 2cos()sin 2sin(),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
9、,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,技能提升练,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,拓展冲刺练,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,16.设，0，且满足sin cos cos sin 1，求sin(2)sin(2)的取值范围.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解 由sin cos cos sin 1，得sin()1，,
