?34?sin70sin30?,?tan(A?B)?0012sin70?034
?1,
2.证明:?A?B??4tanA?tanB1?tanAtanB 得tanA?taBn?taBn??1tAantAanBt aBta?n
A? 1?tan ?(1?taAn3.解:原式?log2(cos?9)?(1Bta?n )cos2?9cos4?9),
而cos?9cos2?918cos4?9sin??9cos?9cos2?9cos4?9?1
8sin?9即原式?log2??3
4.解:f(x)?a?(1)2k??[k??1?cos2x2?a?12sin2x?b?2a2sin(2x??4)?a2?b
?2?2x?,k???4?2k???2,k??3?8?x?k???8,
3?8?8],k?Z为所求
(2)0?x???2,422?2x??4?5?4,?22?sin(2x??4)?1,
f(x)min?1?a?b?3,f(x)max?b?4,
?a?2?22b,? 4数学4(必修)第三章 三角恒等变换 [提高训练C组]
一、选择题 1.C
cos10?sin100020240cos35(cos10?sin10)?cos10?sin10cos35000?2sin55cos3500?2
2.C y?2cos(12?6?x)?cos(?6?x)?cos(?6?x)??1
3.B y?sin2x?32(1?cos2x)?3?12sin2x?32cos2x?32
46
?sin(2x??3)?32,令2x??3?k?,x?k?2??6,当k?2,x?5?6
4.D y?sin2A?2sinB?sin2A?2cosA?1?cos2A?2cosA ??(cosA?1)2?2,而0?cosA?1,自变量取不到端点值
5.C (1?tan210)(1?tan240)?2,(1?tan220)(1?tan230)?2,更一般的结论 ????450,(1?6.A f(x)?1tanx?tanx2ta?n)?(1?ta?n )2?1?(tanx?12)?214,当tanx?12时,f(x)min?4
二、填空题
1. ③ 对于①,sinx?cosx?2sin(x??4)?2?32;
对于②,反例为??300,???3300,虽然???,但是cos??cos?
2?y? 对于③,y?sinxsinx2?(?4?)xs?in(2
2?)2.? y?1?coxssinxxtan591359223.? (sin??cos?)?(sin??cos?)?,2sin(???)??
723636???sixn1cxos??sxin
14.1 y?2sin(x??3),?32?x??3?5?6,ymi?2sinn5?6?1
aco?xs 22?s5.1,?22 y?acosx?bsinxcoxb2asin?x22 ?a?b222sin(2x??)?a2,
a?b222?a2?2,?a?b222?a2??1,a?1,b??22 三、解答题
1. 解:(1)当??0时,f(x)?sinx?cosx? 2k????x?2sin(x??4)
?3???,2k???x?2k??f,(x)为递增;
24244??3??5?,2k???x?2k??,f(x)为递减 2k???x??2k??24244?2k?? ?f(x)为递增区间为[2k?? f(x)为递减区间为[2k??3?4,k2??,k2???45?4k]?,Z; k]?,Z。
?4 47
(2)f(x)? ???k???42cos(x?,k?Z
?4??)为偶函数,则???4?k?
2.解:2(2cos2B?1)?8cosB?5?0,4cos2B?8cosB?5?0 ??a?b3413 得cosB?,sinB?,cos??????,sin??,
5522a?b4?3Bco?s?sin103 sinB(???)3.解:?(?4sBin?c?os?2
513?x)?(?4?x)?,?cos(?4?x)?sin(?4?x)?, ?4?x)?120169 而cos2x?sin(?2?2x)?sin2(?4?x)?2sin(?4?x)cos(
120 ?12。 ?169??513cos(?x)41312a2asin2x?3a23a2(1?cos2x)?32a?b
cos2x4.解:f(x)? ?sinx2?coxs?2b?asxi?n(2?b )3,k??5?12?x?k??11?12? (1)2k?? ?[k???25?12?2x?,k???31?1?2k??],k?3?2
12Z为所求
(2)0?x??2,??3?2x??3?2?3,?32?sin(2x??3)?1
f(x)min??32a?b??2,f(xm)ax?a?b? 3,?3?a?2?a?b??2???? ?2??b??2??a?b?3?3
48