高考复习专题——三角函数综合

高考复习专题——三角函数综合内容摘要：

2232.22223,0).223s i n (2c oss i n:02c oss i n BCCBCBCBCBCCBCB或即或、由得由[法二 ] .4:),0(.s i nc os,0s i n0)c os( s i ns i n.0s i nc osc oss i nc oss i ns i ns i n0)s i n (c oss i ns i ns i n:0s i n)c oss i n(s i n 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( c otc ot )1( .43c os,,, 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] CACACABacbBBt an1t an1c otc ot.s i ns i ns i n:,47)43(1s i n:43c os ( 1) 222于是及正弦定理得由得由[解析 ] .774s i n1s i ns i ns i n)s i n (s i ns i ns i nc osc oss i ns i nc oss i nc os22BBBBCACAACACCCAA.774s i n1s i ns i ns i n)s i n (s i ns i ns i nc osc oss i ns i nc oss i nc os22BBBBCACAACACCCAA.2,2:,43c os,23c os:23 )2( 2bcaBBcaBCBA即可得由得由.3,945 2)(.5c os2:c os2:222222222caaccacaBacbcaBaccab得由余弦定理.)82c o s (,528),2,(),c o s,s i n2()s i n,( c o s 的值求且和已知向量nmnm[例 4] .)82c o s (,528),2,(),c o s,s i n2()s i n,( c o s 的值求且和已知向量nmnm[例 4] 22)s i n( c os)2s i n( c os),s i nc o s,2s i n( c os nmnm[解析 ] .257)4c os (:528)4c os (12)4c os (44)s i n( c os224。