高考數(shù)學(xué)大一輪復(fù)習(xí)第四章三角函數(shù)、解三角形4.5簡單的三角恒等變換第2課時課件.ppt

1、第2課時簡單的三角恒等變換,4.5簡單的三角恒等變換,課時作業(yè),題型分類深度剖析,內(nèi)容索引,題型分類深度剖析,題型一三角函數(shù)式的化簡,自主演練,答案,解析,4sin ,解析,答案,解析,答案,解析,答案,為第二象限角，,(1)三角函數(shù)式的化簡要遵循“三看”原則：一看角，二看名，三看式子結(jié)構(gòu)與特征. (2)三角函數(shù)式的化簡要注意觀察條件中角之間的聯(lián)系(和、差、倍、互余、互補等)，尋找式子和三角函數(shù)公式之間的共同點.,典例 (1)(2018太原質(zhì)檢)2sin 50sin 10(1 tan 10) .,題型二三角函數(shù)的求值,多維探究,命題點1給角求值與給值求值,答案,解析,答案,解析,答案,解析,c

2、os cos() cos()cos sin()sin ,命題點2給值求角,解析,答案,答案,解析,解析tan tan(),答案,解析,cos()cos cos sin sin ,(1)給角求值與給值求值問題的關(guān)鍵在“變角”，通過角之間的聯(lián)系尋找轉(zhuǎn)化方法. (2)給值求角問題：先求角的某一三角函數(shù)值，再求角的范圍確定角.,答案,解析,則(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)已知函數(shù)f(x)sin2xcos2x2 sin xcos

3、 x(xR). (1)求 的值；,題型三三角恒等變換的應(yīng)用,師生共研,解答,解由cos 2xcos2xsin2x與sin 2x2sin xcos x，得,所以f(x)的最小正周期是. 由正弦函數(shù)的性質(zhì)，得,(2)求f(x)的最小正周期及單調(diào)遞增區(qū)間.,解答,三角恒等變換的應(yīng)用策略 (1)進行三角恒等變換要抓住：變角、變函數(shù)名稱、變結(jié)構(gòu)，尤其是角之間的關(guān)系；注意公式的逆用和變形使用. (2)把形如yasin xbcos x化為y sin(x)，可進一步研究函數(shù)的周期性、單調(diào)性、最值與對稱性.,答案,解析,跟蹤訓(xùn)練(1)函數(shù)f(x)sin(x)2sin cos x的最大值為 .,1,解析因為f(x

4、)sin(x)2sin cos x sin xcos cos xsin sin(x)， 又1sin(x)1， 所以f(x)的最大值為1.,解析,答案,化歸思想和整體代換思想在三角函數(shù)中的應(yīng)用,思想方法,思想方法指導(dǎo),思想方法指導(dǎo)(1)討論形如yasin xbcos x型函數(shù)的性質(zhì)，一律化成y sin(x)型的函數(shù). (2)研究yAsin(x)型函數(shù)的最值、單調(diào)性，可將x視為一個整體，換元后結(jié)合ysin x的圖象解決.,規(guī)范解答,規(guī)范解答,課時作業(yè),基礎(chǔ)保分練,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

5、,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,

6、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，結(jié)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

7、,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.已知函數(shù)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

8、,7,8,9,10,11,12,13,14,15,16,技能提升練,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 Btan C1 ， 則角A的值為 .,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,拓展沖刺練,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,16.(2018泉州模擬)已知角的頂點在坐標原點，始邊與x軸的正半軸重合，終邊經(jīng)過點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 2