2019届全国通用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一课时讲义.ppt

,剖析题型 提炼方法,实验解读,构建知识网络 强化答题语句,探究高考 明确考向,4.5,简单的三角恒等变换,第四章三角函数、解三角形,基础知识自主学习,课时作业,题型分类深度剖析,内容索引,基础知识自主学习,1.,两角和与差的余弦、正弦、正切公式,cos(,),cos,cos,sin,sin,(C,(,),),cos(,),(,C,(,),),sin(,),(,S,(,),),sin(,),(,S,(,),),知识梳理,cos,cos,sin,sin,sin,cos,cos,sin,sin,cos,cos,sin,2.,二倍角公式,sin 2,；,cos,2,；,tan 2,.,2sin,cos,cos,2,sin,2,2cos,2,1,1,2sin,2,2.,升幂公式：,1,cos,2,2cos,2,，,1,cos,2,2sin,2,.,【,知识拓展,】,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,.(,),基础自测,1,2,3,4,5,6,7,(3),y,3sin,x,4cos,x,的最大值是,7.(,),(4),公式,tan(,),可以变形为,tan,tan,tan(,)(1,tan,tan,),，且对任意角,，,都成立,.(,),题组二教材改编,1,2,4,5,6,答案,3,2.P127T2,若,cos,，,是第三象限的角，则,等于,解析,7,解析,是第三象限角，,1,2,4,5,6,解析,3.P131T5sin 347cos 148,sin 77cos 58,.,解析,sin 347cos 148,sin 77cos 58,sin(270,77),cos,(90,58),sin 77cos 58,(,cos,77)(,sin 58),sin 77cos 58,sin 58cos 77,cos,58sin 77,3,7,答案,4.P146T4tan 20,tan 40,tan 20tan 40,.,解析,1,2,4,5,6,3,答案,7,tan 20,tan 40,tan 60(1,tan 20tan 40),5.,化简：,.,解析,1,2,4,5,6,3,答案,7,题组三易错自纠,6.(2018,昆明模拟,),若,tan,，,tan(,),，则,tan,.,解析,1,2,4,5,6,3,7,答案,解析,1,2,4,5,6,3,7,答案,1,2,4,5,6,3,7,1,2,4,5,6,3,7,题型分类深度剖析,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,答案,解析,解析,答案,3.,计算,的值为,.,解析,答案,(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,.,思维升华,典例,(1),设,，,都是锐角，且,cos,，,sin(,),，则,cos,.,题型二和差公式的灵活应用,多维探究,命题点,1,角的变换,答案,解析,于是,cos,cos,(,),cos,(,),cos,sin(,)sin,(2)(2017,泰安模拟,),已知,cos,(75,),，则,cos,(30,2,),的值为,.,答案,解析,命题点,2,三角函数式的变换,解答,解答,引申探究,解答,(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,.,思维升华,跟踪训练,(1)(2017,豫北名校联考,),计算：,.(,用数字作答,),答案,解析,答案,解析,用联系的观点进行三角变换,思想方法,思想方法指导,2,思想方法指导,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,.,答案,解析,(2),原式,1,tan 17,tan 28,tan 17tan 28,1,tan 45(1,tan 17tan 28),tan 17tan 28,1,1,2.,cos,sin,，,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,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,3.(2017,西安二检,),已知,是第二象限角，且,tan,，则,sin 2,等于,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,4.(2017,河南洛阳一模,),设,a,cos,50cos 127,cos,40sin 127,，,b,(sin 56,cos,56),，,c,，则,a,，,b,，,c,的大小关系是,A.,a,b,c,B.,b,a,c,C.,c,a,b,D.,a,c,b,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,a,sin 40cos 127,cos,40sin 127,sin(40,127),sin 167,sin 13,，,sin(56,45),sin 11,，,sin 13sin 12sin 11,，,a,c,b,.,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,答案,解析,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,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,6tan,6,1,tan,(tan,1),，,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,11.,已知,sin,cos,，则,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,12.(2018,吉林模拟,),已知,sin(,),cos,cos,(,)sin,，,是第三象,限角，则,.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,解析,依题意可将已知条件变形为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,技能提升练,答案,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,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,15.(2017,武汉调研,),设,，,0,，,，且满足,sin,cos,cos,sin,1,，,则,sin(2,),sin(,2,),的取值范围为,.,拓展冲刺练,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,1,1,解析,由,sin,cos,cos,sin,1,，得,sin(,),1,，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,sin(2,),sin(,2,),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,16.(2017,合肥模拟,),已知函数,f,(,x,),(2cos,2,x,1)sin 2,x,cos,4,x,.,(1),求,f,(,x,),的最小正周期及单调递减区间；,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,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,