半角公式及万能公式

半角的正弦、余弦、正切及万能公式2221122sincoscos一、复习：22122122coscos,cossin212212222coscos,cossin得：换成将212212coscos,cossin二、半角公式：coscostan112212212coscoscossin半角公式一：.所在象限确定符号由2222cossintan半角公式二：222222cossinsinsincos1222cossintan222222coscossincossin1cossinsincostan112cossincossin11cossinsincostan112半角公式二：sincoscossincot11222cottan化简：sinsincossin21cos1csc2sincos1)cos()sin(212)tan(24cossincossin11cossincossin11.tan,tan,cos,sin),(,cos的值求：已知二、例422202531.cos,sin),(),(020204202解：,cossin55212552212coscos,cossintan212222214sincostan5251521.cos,)cos()sin(值求：已知例xxx441443241424)cos(])sin[(xx解：4142)(cosx.)(cos4142x即：222142)cos()(cosxx又221xsin41212xsinxx22142sincos2121212)()tantan)(tan(cot21223：化简：例)sincoscossin)(sincossincos(1111cossincos12sin2csc2三、万能公式利用二倍角公式推导：.1222cossinsin.tan,cos,sintan表示提出问题：试用22222222cossincossin21222tantan2222sincoscos22222222sincossincos212122tantan21222tantantan三、万能公式,tantansin21222,tantancos21212221222tantantan则若令,tan2t.tan,cos,sin2222121112tttttt这样“三角”与“代数”沟通起来，因此称为“万能公式”。弦化切的两种方法：“齐次式”弦化切及万能公式.8182tantan练习：421sin428182tantan421tan21)(tan)(tan414122)cos(222sin.cos,cossincossin值求：已知例2111122222222222222222222111122cot)cos(sinsin)sin(coscoscossinsincossincossin)cos(sin)cos(cossincossin解：5321121121212222)()(tantancos.tan,cossin的值求且：已知例45404505322200592212)cos(sin：解法54sin21222tantan22212tantan或2212270222500tantan41422tantan.,tan251251414135451120tan..tan2514.tan,cossin的值求且：已知例45404505322200592212)cos(sin：解法,sin54)sin)(cossin(cos2222)sin(cos22535122sincos5225125322sincossincos.sincostan251221453cos,00540450又.tancossin,)tan(12221432求：已知例,tantantan)tan(321114解：tan1)2cos1(tan1tan2tan1cos22sin22tan1)tan1tan11(tan1tan222252cossin)cos(sincoscossincoscossin222122原式法2:1122tantancos101322tansectan1012cos5213131012原式例4：已知的值求)4sin(21sin2cos2),,2(2,222tan2解：)4sin(2sincos)4sin(21sin2cos22tan1tan1,222tan22tan2tan22tan1tan22或即2tan)2,4(),2(2sincossincos22302125222sincos,tantan求证：：已知例2222112sintantansincos证：2221sintantan222121121sintan)tan(222221sincossincossin22sinsin0.原式得证tantantan:,sin)tan(2226求证满足：、：已知锐角例cossintantantantan:21由已知得：证tancossintancossintan2122212sincossincossin)sin(cos)cos(sin222112)sin(coscos)sincos(sin222232213tantantan2221331422tantantantantantantantan22133tantantantantantan22故：成立。tantantan22tantantan:,sin)tan(2226求证满足：、：已知锐角例2121tantantantantantan证：由已知得：）（10332tantantantantantantantantan,tantantan:21422只需证：欲证0314322tantantantantan)tan)(tantan(tan即证：即证：.)(问题得证已证，而此式就是1.cos,)sin(,tan),,(值求、例：若BBAABA1352120,tantancos,tantansin532121542122222AAAAAA解：),(),sin(sin013554BBAA又为锐角，AAA13554BA1312)cos(BA6516ABAABAABABsin)sin(cos)cos(])cos[(cos220222312322求证：为锐角，、例：已知,sinsinsinsin22232213sin1)22sinsincossin(由条件得：法解：222sinsincoscos)cos(22332sinsinsincos03322cossinsincos),(230222220222312322求证：为锐角，、例：已知,sinsinsinsin22232213sin2)22sinsincossin(由条件得：法解：2cottan:两式相除得)tan(tan22)(Zkk2222k),(230222例7：已知，)4,0(),43,4(,135)4cos(,53)4sin(且)sin(求解:)](2cos[)sin()]4()4cos[()]4sin()4sin()4cos()4[cos(54)4cos()43,4(,53)4sin(且1312)4sin(),4,0(,135)4cos(且6556)13125313554(上式应用：找出已知角与未知角之间的关系