2022年同角三角函数基本关系教案人教A版数学必修四第一章 .pdf

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1、学习必备欢迎下载第一章三角函数1.2 任意角的三角函数1.2 1 同角三角函数的基本关系课型：新授课课时：第一课时1、教学目标1.知识与技能目标：通过观察猜想出两个公式，运用数形结合的思想让学生掌握公式的推导过程，理解同角三角函数的基本关系式，掌握基本关系式在两个方面的应用：1）已知一个角的一个三角函数值能求这个角的其他三角函数值；2）证明简单的三角恒等式。2.过程与方法：培养学生观察猜想证明的科学思维方式；通过公式的推导过程培养学生用旧知识解决新问题的思想；通过求值、证明来培养学生逻辑推理能力；通过例题与练习提高学生动手能力、分析问题解决问题的能力以及其知识迁移能力。3.情感、态度与价值观：

2、经历数学研究的过程，体验探索的乐趣，增强学习数学的兴趣。2、教学重点和难点重点：同角三角函数基本关系式的推导及应用。难点：同角三角函数函数基本关系在解题中的灵活选取及使用公式时由函数值正、负号的选取而导致的角的范围的讨论。3、专家建议在公式的推导中，教师是用创设问题的形式引导学生去发现关系式，多让学生动手去计算，体现了 教师为引导,学生为主体,体验为红线,探索得材料,研究获本质,思维促发展 的教学思想；学习必备欢迎下载通过两种不同的例题的对比，让学生能够明白到关系式中的开方，是需要考虑正负号，而正负号是与角的象限有关，角的象限题目可以直接给出来，但有时是需要已知条件来推出角可能所在的象限，通过

3、分析，把本节课的教学难点解决了；课堂在完成例题及变式时要给予学生充分的时间思考与尝试，故对学生的检测只能安排在课后的作业中，作业可以检测学生对本节课内容掌握的情况，能否灵活运用知识进行合理的迁移，可以发现学生在解题中存在的问题4、教法与学法1教法：采取诱思探究性教学方法，在教学中提出问题，创设情景引导学生主动观察、思考、类比、讨论、总结、证明，让学生做学习的主人，在主动探究中汲取知识，提高能力。2学法：从学生原有的知识和能力出发,在教师的带领下，通过合作交流,共同探索,逐步解决问题.数学学习必须注重概念、原理、公式、法则的形成过程，突出数学本质。5、教具：计算机，多媒体投影，黑板6、教学过程设

4、计6.1 创设情境引入课题【师】我们看下面一个问题【板书/PPT】计算下列各式的值答案：1,1，33，3322(1)sin 30cos 30;22(2)sin 90cos 90;sin30(3);cos30sin 60(4).cos60文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7

5、T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档

6、编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7

7、T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档

8、编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7

9、T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档

10、编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3文档编码:CV4U5H2H3P4 HL3K7T5O9H6 ZM3P3G9Q9B3学习必备欢迎下载【师】思考：【板书/PPT】问题 1：从以上的过程中，你能发现什么一般规律？问题 2：你能否用代数式表示这两

11、个规律？6.2 自主学习推导公式【师】我们证明一下【板书/PPT】1证明公式：（同角三角函数基本关系）（1）、平方关系:1cossin22（2）、商的关系:tancossin回忆：任意角三角函数的定义？学生回答：设 是一个任意角，它的终边与单位圆交于点P（x，y）则：sin=y；cos=x，xytan引导学生注意：单位圆中122yx所以：sin 2+cos2=122xy；cossin=tanxy由三角函数的定义，我们可以得到以下关系：（1）商数关系：consintan（2）平方关系：1sin22con说明：注意“同角”，至于角的形式无关重要，如22sin 4cos 41等；注意这些关系式都是对

12、于使它们有意义的角而言的，如tancot1(,)2kkZ；对这些关系式不仅要牢固掌握，还要能灵活运用（正用、反用、变形用），如：2cos1 sin，22sin1cos，sincostan等。文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P

13、8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文

14、档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW

15、1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U

16、9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 H

17、W9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R

18、4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1

19、ZO5E3P8R1H2学习必备欢迎下载 6.3 小组合作及时训练一、求值问题【师】我们看一下例1【板书/PPT】例 1已知3sin-5，并且是第三象限角，求 cos,tan变式 1、已知12cos13，求sin,tan解：（1）22sincos1，222234cos1sin1()()55又是第三象限角，cos0，即有4cos5，从而sin3tancos4【师】我们看一下变式【板书/PPT】变式 122sincos1，2222125sin1cos1()()1313，又12cos013，在第一或四象限角。当在第一象限时，即有sin0，从而3sin13，sin5tancos12；当在第四象限时，即有

20、sin0，从而5sin13，sin5tan-cos12【师】我们看一下变式【板书/PPT】变式 2.已知 tan0m，求sin和cos解：22sincos1，sintancos，2222(costan)coscos(1tan)1，即有221cos1tan，又tan为非零实数，为象限角。文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1

21、ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P

22、8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文

23、档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW

24、1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U

25、9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 H

26、W9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R

27、4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2学习必备欢迎下载当在第一、四象限时，即有cos0，从而22211cos11mmm，221sintancos1mmm；当在第二、三象限时，即有cos0，从而22211cos11mmm，221sintancos1mmm总结：1.已知一个角的某一个三角函数值，便可运用基本关系式求出其它三角函数值。在求值中，确定角的终边位置是关键和必要的。有时，由于角的终边位置的不确定，因此解的情况不止一种。2.解题时产生遗漏的主要原因是：没有确定好或不去确定角的终边位置；利用平方关系开平方时，漏掉了

28、负的平方根。【师】我们看一下例2【板书/PPT】例 2、已知cos2sin，求（1）cos2sin5cos4sin222sin2sincoscos（2）解：（1）2tancos2sin611222tan54tancos2sin5cos4sin222222222sin2sincoscos2sin2sincoscos=sin+cos2tan2tan1=tan135（2）强调（指出）技巧：1分子、分母是正余弦的一次（或二次）齐次式注意所求值式的分子、分母均为一次齐次式，把分子、分母同除以cos,将分子、分母转化为tan的代数式；2“化 1 法”可利用平方关系1cossin22，将分子、分母都变为二次

29、齐次式，再利用商数关系化归文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E

30、3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H

31、2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:

32、CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q

33、8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10

34、 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T1

35、0R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2学习必备欢迎下载为 tan的分式求值；二、化简【师】我们看一下例3【板书/PPT】例 3化简21 cos 620 cos620cos(3602

36、60)解因为cos260cos(18080)cos80【师】我们看一下变式【板书/PPT】变式、化简22sin1 coscos1sincos0解因为sin|sin|,|cos|cos所以 原式小结：化简三角函数式，化简的一般要求是：（1）尽量使函数种类最少，项数最少，次数最低；（2）尽量使分母不含三角函数式；（3）根式内的三角函数式尽量开出来；（4）能求得数值的应计算出来，其次要注意在三角函数式变形时，常将式子中的“1”作巧妙的变形，三、证明恒等式【师】我们看一下例4【板书/PPT】例 4求证：cos1sin1sincosxxxx证法一：由题义知cos0 x，所以1sin0,1sin0 xx左

37、边=2cos(1sin)cos(1sin)(1sin)(1sin)cosxxxxxxx1sincosxx右边原式成立2 tan,0,2 tan,0,22;2kk当22;2kk当322;2kk当3222;2kk当kZ文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R

38、4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1

39、ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P

40、8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文

41、档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW

42、1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U

43、9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 H

44、W9T10R4W4G1 ZO5E3P8R1H2学习必备欢迎下载证法二：由题义知cos0 x，所以1sin0,1sin0 xx又22(1sin)(1sin)1sincoscoscosxxxxxx，cos1sin1sincosxxxx证法三：由题义知cos0 x，所以1sin0,1sin0 xxcos1sin1 sincosxxxxcoscos(1sin)(1sin)(1sin)cosxxxxxx22cos1sin0(1sin)cosxxxxcos1sin1sincosxxxx总结：证明恒等式的过程就是分析、转化、消去等式两边差异来促成统一的过程，证明时常用的方法有：（1）从一边开始，证明它等于另

45、一边；（2）证明左右两边同等于同一个式子；（3）证明与原式等价的另一个式子成立，从而推出原式成立。题型四利用 sin cos 与 sin cos 的关系【师】我们看一下例5【板书/PPT】例 5 已知 0，sin cos，求 tan 的解：由 sin cos 15得 sin cos 12250，又 00，cos 0，则 sin cos 0，化简：sin 1cos tan sin tan sin 解tan sin tan sin sin cos sin sin cos sin sin sin cos sin sin cos 1cos 1cos 1cos 21cos 1cos 1cos 2sin2

46、1cos sin 原式sin 1cos 1cos sin 1.2.若 sin A45，且 A是三角形的一个内角，求5sin A815cos A7的值解：sin A450，A为锐角或钝角，当 A为锐角时，cos A1sin2A35，原式 6.当A为钝角时，cos A1sin2A35，原式545815 35734.3.已知tan2,180270，求3sin12cos文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5

47、E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1

48、H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码

49、:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2

50、Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K10 HW9T10R4W4G1 ZO5E3P8R1H2文档编码:CW1K2Q8U9K1