高三数学第二轮冲刺课件(专题五三角函数)全国通用 课件VIP免费

三角函数综合第一课时：三角变换第一课时：三角变换[课前导引]第一课时：三角变换)(coscos,3tantan,3.1则设23D.233C.63B.61A.[课前导引][解析].6333sincoscos,3coscos3sincoscos)sin(coscoscossincossincossincossintantan).24cot(,sin12cos:.2求已知,312tan12tan,2sin32cos02sin2cos)2sin2(cos)2sin22(cos,sin12cos222或或).24cot(,sin12cos:.2求已知[法一].2)24cot(0)24cot(,2tan12tan1)24tan(1)24cot(或.2)24cot(0)24cot(,2tan12tan1)24tan(1)24cot(或[法二]),2cos(1)2sin(2,sin12cos.2)24cot(0)24cot()24cos()24sin(20)24cos(),24(cos2)24cos()24sin(42或或[链接高考][链接高考])(sincoscos41cossin33sinsin021cottan1:,sin2tan,)1(1.22222其中正确的是给出以下四个论断已知中在CBABABABACBAABC[例1]32D.41C.42B.31A.[链接高考])(sincoscos41cossin33sinsin021cottan1:,sin2tan,)1(1.22222其中正确的是给出以下四个论断已知中在CBABABABACBAABC[例1]32D.41C.42B.31A.B)(,cossin2sin1,20)2(则且设xxxx232D.454C.474B.0A.xxxx)(,cossin2sin1,20)2(则且设xxxx232D.454C.474B.0A.xxxxC)(,)3(不等关系中正确的是下列、对任意的锐角coscos)cos(D.sinsin)cos(C.coscos)sin(B.sinsin)sin(A.)(,)3(不等关系中正确的是下列、对任意的锐角coscos)cos(D.sinsin)cos(C.coscos)sin(B.sinsin)sin(A.D)()(,),(')(',),(')(,sin)()4(20051010xfNnxfxfxfxfxxfnn则设xxxxcosD.cosC.sinB.sinA.)()(,),(')(',),(')(,sin)()4(20051010xfNnxfxfxfxfxxfnn则设xxxxcosD.cosC.sinB.sinA.C._________cos,)45()1cos()5(3425则的系数相等的展开式中的系数与中的展开式已知xxxx._________cos,)45()1cos()5(3425则的系数相等的展开式中的系数与中的展开式已知xxxx22.,02cossin,0sin)cos(sinsin,的大小、、求角中已知在CBACBCBBAABC[例2].,02cossin,0sin)cos(sinsin,的大小、、求角中已知在CBACBCBBAABC[例2].0)cos(sinsin.0sincoscossincossinsinsin,0)sin(cossinsinsin:,0sin)cos(sinsinAABBABABABABABABACBBA即得由[法一]0)43(2cossin:02cossin.43.4:),0(.sincos,0sin),,0(BBCBCBAAAABB得由从而知由从而.125,3,4.125,3,21cos:.0cossin2sin.02sinsinCBACBBBBBBB由此得亦即即.22232.22223,0).223sin(2cossin:02cossinBCCBCBCBCBCCBCB或即或、由得由[法二].4:),0(.sincos,0sin0)cos(sinsin.0sincoscossincossinsinsin0)sin(cossinsinsin:0sin)cossin(sinAAAABAABBABABABABABABACBBA知由即得由.125,3,4.125,3:,212.252,43CBACBBCCBCB得再由不合要求知从而.,23)2(cotcot)1(.43cos,,,,,,,,,的值求设的值；求成等比数列已知分别为的对边内角中caBCBACABcbacbaCBAABC[例3]CACACABacbBBtan1tan1cotcot.sinsinsin:,47)43(1sin:43cos(1)222于是及正弦定理得由得由[解析].774sin1sinsinsin)sin(sinsinsincoscossinsincossincos22BBBBCACAACACCCAA.774sin1sinsinsin)sin(sinsinsincoscossins...