(通用版)2020版高考数学大一轮复习第21讲两角和与差的正弦学案理新人教A版.docx

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1、1第 21讲 两角和与差的正弦、余弦和正切两角和与差的正弦、余弦、正切公式(1)公式 S( ):sin( )= . (2)公式 C( ):cos( )= . (3)公式 T( ):tan( )= . 常用结论1.两角和与差的正切公式的变形:tan tan = tan( )(1tan tan ).2.二倍角余弦公式的变形:sin2= ,cos2= .1-cos22 1+cos223.一般地,函数 f( )=asin +b cos (a,b为常数)可以化为 f( )= sin(+ )a2+b2或 f( )= cos(- ) .(其中 tan =ba) a2+b2 (其中 tan =ab)题组一 常

2、识题1.教材改编 sin 75的值为 . 2.教材改编 已知 cos =- , ,则 sin + 的值是 . 35 (2, ) 33.教材改编 cos 65cos 115-cos 25sin 115 = . 4.教材改编 已知 tan = ,tan =- 2,则 tan(- )的值为 . 13题组二 常错题索引:忽略角的取值范围;公式的结构套用错误;混淆两角和与差的正切公式中分子、分母上的符号;方法选择不当致误 .5.已知 tan + = , , ,则 cos 的值是 . 54 17 226.化简: sin x- cos x= . 12 327.计算: = . 1-tan151+tan158.

3、若 += ,则1 +tan( - )(1-tan )的值为 . 34探究点一 两角和与差的三角函数公式例 1 (1)2018湘潭模拟 若 sin(2- )= ,sin(2+ )= ,则 sin 2 cos = ( )16 12A. B.23 13C. D.16 112(2)2018晋城一模 已知 cos = cos ,tan = ,则 tan(+ )= ( +6) 3 33. 总结反思 两角和与差的三角函数公式可看作是诱导公式的推广,可用 , 的三角函数表示 的三角函数,在使用两角和与差的三角函数公式时,特别要注意角与角之间的关系,完成统一角和角与角转换的目的 .变式题 (1)2018佛山质检

4、 已知 cos = , ,则 cos = ( )17 (0,2) ( -3)A.- B.1114 3314C. D.5314 1314(2)2018唐山三模 已知 tan + =1,则 tan - = ( )6 6A.2- 3B.2+ 3C.-2- 33D.-2+ 3探究点二 两角和与差公式的逆用与变形例 2 (1)2018烟台一模 已知 cos = ,则 cos x+cos = ( )(x-6) 33 (x-3)A.-1 B.1C. D.233 3(2)已知 sin + cos = ,sin - cos = ,则 sin(- )= . 13 12总结反思 常见的公式变形:(1)两角正切的和差

5、公式的变形,即 tan tan = tan( )(1tan tan );(2) asin +b cos = sin(+ ) tan =a2+b2ba.变式题 (1)2018河南中原名校联考 cos 375+ sin 375的值为 ( )22 22A. B. C.- D.-32 12 32 12(2)(1+tan 20)(1+tan 21)(1+tan 24)(1+tan 25)= . 探究点三 角的变换问题例 3 (1)已知 ,cos -sin = ,则 sin + 的值是 ( )(-3,0) ( +6) 435 12A.- B.-235 210C. D.-235 45(2)2018莆田二模

6、已知 sin = ,sin(- )=- , , 均为锐角,则 = ( )255 1010A. B. C. D.512 3 4 64总结反思 常见的角变换: 2= 2 ,2= (+ )+(- ),2 4= + , += - - 等 . +2 -2 3 2 6变式题 (1)2018榆林模拟 若 0 0, tan(- )= = = .tan -tan1+tan tan 2tan1+3tan2 21tan +3tan tan 0, +3tan 2 =2 , tan(- ) ,当且仅当 3tan2= 1,即 tan 1tan 1tan 3tan 3 33= 时取等号,此时 = ,tan = 3tan ,

7、即 tan = ,= .33 6 3 3又 0 , 0- ,2 2 0tan(- ) ,33又 y=tan x在 上单调递增,(0,2) 当 tan(- )取得最大值时, - 的值最大, 当 = ,= 时, - 的值最大, -3 6 的最大值为 - = .366例 2 配合例 3使用 2018安徽皖江八校联考 的值为 . 2cos55- 3sin5cos5答案 1解析 = = =1.2cos55- 3sin5cos5 2cos(60-5)- 3sin5cos5 cos5+ 3sin5- 3sin5cos5例 3 配合例 3使用 2018安阳模拟 已知 m= ,若 sin 2(+ )=3sin tan( + + )tan( - + )2 ,则 m= ( )A. B.12 34C. D.232解析 D sin 2(+ )=3sin 2 , sin(+ )+(+- )=3sin(+ )-(+- ), sin(+ )cos(+- )+cos(+ )sin(+- )=3sin(+ )cos(+- )-3cos(+ )sin(+- ),- 2sin(+ )cos(+- )=-4cos(+ )sin(+- ),9即 = =2,sin( + + )cos( + - )cos( + + )sin( + - )tan( + + )tan( - + )m= 2.故选 D.

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