高考復(fù)習(xí)專題——三角函數(shù)綜合-資料下載頁(yè)
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【導(dǎo)讀】下列、對(duì)任意的銳角??
【正文】 .)(),()23s i n (32)2316c os ()2316c os ()( 最小正周期的值域和并求函數(shù)化簡(jiǎn)xfZkRxxxkxkxf?????????????[例 2] .)(),()23s i n (32)2316c os ()2316c os ()( 最小正周期的值域和并求函數(shù)化簡(jiǎn)xfZkRxxxkxkxf?????????????[例 2] [解析 ] )23s i n (32)2316c os ()2316c os ()( xxkxkxf???????????.2)(]4,4[)(.2c os4)23s i n (32)23c os (2?????????????Txfxfxxx的周期函數(shù)；的值域?yàn)楹瘮?shù).,)()2)(,(2s i n2 )2( 。],3,3[31)( )1( .),2s i n3,( c os),1,c os2(,)( 的值、求實(shí)數(shù)的圖象平移后得到函數(shù)量的圖象按向若函數(shù)求且若其中向量設(shè)函數(shù)nmxfymnmcxyxxxfRxxxbxabaxf???????????????????[例 3] .23)62s i n ( :31)62s i n (21).62s i n (212s i n3c o s2)(: 1)( 2???????????????xxxxxxf得由依題設(shè)[解析 ] .4???x即,362,65622,33???????????????????xxx?.)(2s i n2),(2s i n2 2)( 的圖象平移后得到函數(shù)向量的圖象按函數(shù)nmxynmcxy??????1 ,12 ,21)12(2s i n2)( )1(????????nmmxxf????得由? ?.,。,))()(39。( 0)(39。)(,0)() ) ,42t an (),42s i n (2() ) ,42t an (,2c os2( 則證明之不存在若的值則求出？若存在的導(dǎo)函數(shù)是其中使是否存在實(shí)數(shù)令已知向量xxfxfxfxfxbaxfxxbxxa??????????????????[例 4] [解析 ] 2t an112t an2t an12t an1 )2c os222s i n22(2c os22)42t an ()42t an ()42s i n (2c os22)( xxxxxxxxxxxbaxf????????????????????.c o ss i n12c o s22c o s2s i n2 2 xxxxx ?????? ?0)(39。)(,02 ,20c os2 s i nc osc oss i n)(39。)( ,0)(39。)(??????????????xfxfxxxxxxxxfxfxfxf使所以存在實(shí)數(shù)可得即：令???
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