1课题:两角和与差的正弦、余弦、正切公式2cos(α–β)=cosαcosβ+sinαsinβ思考cos(α+β)=?将看为为)(3)cos()sin(sin)cos(cos)](cos[sinsincoscos4cos(α+β)=cosαcosβ–sinαsinβ公式的结构特征:左边是复角α+β的余弦,右边是单角α、β的余弦积与正弦积的差.简记:)(C和角的余弦公式)cos(sinsincoscos5cos2cos2sin2sincos2cossincoscossinsin二、公式的推导6)sin(cos)cos(sinsin二、公式的推导)](sin[sincoscossin7两角和与差的正弦公式1、两角和的正弦公式sin)sincoscossin(sin)sincoscossin(2、两角差的正弦公式简记:()S简记:()S8sin)sincoscossin(cos)coscossinsin(9练习:课本131页1(1)(2)15sin)1(30sin45cos30cos45sin)3045sin(426212223221075cos)2(练习:课本131页1(1)(2)30sin45sin30cos45cos)3045cos(4262122232211练习:创新基础测评1,135sin为锐角,∵分析:,1312)135(1sin1cos22,sin6coscos6sin)6sin(13523131221263513612练习:创新基础测评34sin127cos4cos127sin)4127sin(3sin2313练习:创新基础测评231)cos(∵分析:,31sinsincoscos41)cos(∵,41sinsincoscos,127coscos2得:①②①+②247coscos14练习:报纸随堂练习1255sin105sin)105360sin()60sin45cos60cos45(sin)6045sin(462)23222122(15练习:报纸随堂练习5,2,20∵解:)(sin1)cos(2,232,6556)6533(12,1312)135(1cos1sin2216])sin[(sin练习:报纸随堂练习5sin)cos(cos)sin(5313126556)135(653317练习:报纸随堂练习6)sin()sin(BABA∵解:BABABABAsincoscossinsincoscossin,26cossin2BA,46cossinBA)cos()cos(BABA同理:,22coscos2BA,42coscosBA18,34246tancoscoscossinABABA练习:报纸随堂练习6,60A45,222346cosBB,46cos60sinB19两角和的正切公式:sinαcosβ+cosαsinβcosαcosβ-sinαsinβsin(α+β)cos(α+β)coscos0当时,coscos分子分母同时除以tanα+tanβtan(α+β)=1-tanαtanβtan()()记:+T20上式中以代得tanα+tanβtan(α+β)=1-tanαtanβtantan()tan[()]1tantan()tanα-tanβ=1+tanαtanβtanα-tanβ∴tan(α-β)=1+tanαtanβ()记-T21tanαtanβtan(αβ)=1tan++-αtanβ()记:+Ttanαtanβtan(αβ)=1tan--+αtanβ()记:-T注意:必须在定义域范围内使用上述公式。即:tan,tan,tan(±)只要有一个不存在就不能使用这个公式,只能(也只需)用诱导公式来解。如:已知tan=2,求不能用tan()2()T两角和与差的正切公式22遇到这类计算时,怎么办?)2tan()2tan(注意)2cos()2sin(sincostan123tanα+tanβtan(α+β)=1-tanαtanβtanα-tanβtan(α-β)=1+tanαtanβ变形:tanα+tanβ=tan(α+β)(1-tanαtanβ)tanα-tanβ=tan(α-β)(1+tanαtanβ)小结24例3的值。是第四象限角,求已知)4tan(),4cos(),4sin(,53sin是第四象限角,得:解:由,53sin,54)53(1sin1cos22.435453cossintan25于是有例3的值。是第四象限角,求已知)4tan(),4cos(),4sin(,53sin)4sin(;1027)53(225422sin4coscos4sin26)4cos(例3的值。是第四象限角,求已知)4tan(),4cos(),4sin(,53sin;1027)53(225422sin4sincos4cos27)4tan(例3的值。是第四象限角,求已知)4tan(),4cos(),4sin(,53sin7)43(11434tantan14tantantan11tan28练习:课本131页4)4tan(解:231134tantan14tantan练习:创新基础测评429例415tan115tan1)3(70sin20sin70cos20cos)2(42sin72cos42cos72sin)1(算下列各式的值:利用和(差)角公式计42sin72cos42cos72sin1)解:()4272sin(30sin;213070sin20sin70cos20cos)2()7020cos(15tan115tan1)3(90cos015tan45tan115tan45tan15tan1115tan1)1545tan(60tan;331练习:课本131页5(1)(2)(3)18sin72cos18cos72sin)1()1872sin(12sin72sin12cos72cos)2(90sin;1)1272cos(33tan12tan133tan12tan)3(60cos;21)3312tan(45tan132练习:课本132页5(4)(5)14cos74sin14sin74cos)4(74sin14cos74cos14sin)7414sin(26cos34cos26sin34sin)5()60sin(;2326sin34sin26cos34cos)26sin34sin26cos34(cos)2634cos(60cos;2133练习:课本132页5(6)70sin160cos110cos20sin70sin160cos)70180cos()160180sin(70sin160cos)70cos(160sin70sin160cos70cos160sin)70sin160cos70cos160(sin)70160sin(90sin134练习:报纸随堂练习318tan3118tan318tan60tan118tan60tan)1860tan(42tan35练习:报纸随堂练习2,3)tan(∵分析:)]()tan[()22tan()tan()tan(1)tan()tan(433313336练习:课本132页6xxsin23cos21)1(xxsin30coscos30sinxxcossin3)2();30sin(x)cos21sin23(2xx)30sin(2x)cos30sinsin30(cos2xx37xxsin6cos2)4(练习:课本132页6)sin23cos21(22xx)30sin(22x)sin30coscos30(sin22xx38)cos(sin2)3(xx)cos22sin22(2xx)45sin(2x)cos45sinsin45(cos2xx练习:课本132页639练习:课本137页4,1411)cos(,71cos都是锐角,,∵解:,734)71(1cos1sin22,1435)1411(1)(cos1)sin(2240])cos[(cos练习:课本137页4sin)sin(cos)cos(21734143571141141练习:课本137页10的两个实数根,是方程∵解:0732tan,tan2xx,23tantantantan1tantan)tan(,27tantan31)27(12342练习:课本137页12,,:CADBAD依题意,设解21tan,31tan则,tantan1tantan)tan(,12131121314545BAC,即:43练习:课本137页11)]()tan[(2tan)tan()tan(1)tan()tan(;7453153)]()tan[(2tan同理81-44练习:课本137页13(1)(2)(3)xxcos53sin153)1()sin21sin23(56xx)cos30sinsin30(cos56xx)30sin(56x45xxsin23cos23)2(练习:课本137页13(1)(2)(3))sin21cos23(3xx)sin60coscos60(sin3xx)60sin(3x462cos2sin3)3(xx练习:课本137页13(1)(2)(3))2cos212sin23(2xx)2cos30sin2sin30(cos2xx)302sin(2x47练习:课本137页13(9)125tan1125tan45tan)9()1254tan(125tan4tan1125tan4tan32tan33tan练习:课本137