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1、学习好资料欢迎下载1 已知函数 f(x)tan 2x4.(1)求 f(x)的定义域与最小正周期；(2)设 0，4，若 f2 2cos2，求 的大小2 在 ABC 中，a，b，c 分别为内角A，B，C 所对的边长，a3，b2，12cos(BC)0，求边 BC 上的高3 已知 ABC 的一个内角为120，并且三边长构成公差为4 的等差数列，则ABC 的面积为 _4 在 ABC 中，若 b5，B4，tanA2，则 sinA_；a_.5 在 ABC 中，若 b5，B4，sinA13，则 a_.6ABC 的内角 A、B、C 的对边分别为a、b、c，asinAcsinC2asinCbsinB.(1)求 B
2、；(2)若 A75，b2，求 a，c.图 15 7 如图 15，ABC 中，ABAC2，BC23，点 D 在 BC 边上，ADC45，则AD 的长度等于 _2 8 若 ABC 的面积为3，BC2，C60，则边 AB 的长度等于 _9 设 ABC 的内角 A、B、C 所对的边分别为a、b、c，已知 a1，b 2，cosC14.(1)求 ABC 的周长；(2)求 cos(AC)的值10 在 ABC 中，角 A，B，C 的对边分别是a，b，c，已知 sinCcosC1sinC2.(1)求 sinC的值；(2)若 a2b24(ab)8，求边 c 的值11 ABC 的三个内角A，B，C 所对的边分别为a
3、，b，c，asinAsinBbcos2A2a，则ba()A2 3 B2 2 C.3 D.2 12 ABC 的三个内角A，B，C 所对的边分别为a，b，c，asinAsinBbcos2A2a.(1)求ba；(2)若 c2b23a2，求 B.13 ABC 中，B120，AC7，AB5，则 ABC 的面积为 _14 在 ABC 中，内角 A，B，C 的对边分别为a，b，c.已知cosA 2cosCcosB2cab.(1)求sinCsinA的值；(2)若 cosB14，ABC 的周长为 5，求 b 的长15 在 ABC 中，角 A，B，C 所对的边分别为a，b，c.已知 sinAsinCpsinB(p
4、R)，且 ac14b2.(1)当 p54，b 1 时，求 a，c 的值；(2)若角 B 为锐角，求p 的取值范围16 在 ABC 中，角 A，B，C 的对边分别是a，b，c，已知 3acosAccosBbcosC.(1)求cosA 的值；(2)若 a1，cosBcosC2 33，求边 c 的值17 在 ABC 中，内角 A，B，C 的对边分别为a，b，c.已知 BC,2b3a.(1)求 cosA 的值；(2)求 cos 2A4的值18 设 aR，f(x)cosx(asinxcosx)cos22x 满足 f 3f(0)求函数 f(x)在4，1124精品资料-欢迎下载-欢迎下载 名师归纳-第 1
5、页，共 8 页 -学习好资料欢迎下载上的最大值和最小值19 设函数 f(x)sinxcosx3cos(x)cos x(xR)(1)求 f(x)的最小正周期；(2)若函数 yf(x)的图象按 b4，32平移后得到函数yg(x)的图象，求 yg(x)在 0，4上的最大值20 函数 f(x)2cos2x3sin2x(xR)的最小正周期和最大值分别为()A2，3B2，1C，3D，1 21 函数 f(x)Asin(x )A0，0，|2的部分图象如图所示(1)求 f(x)的最小正周期及解析式；(2)设 g(x)f(x)cos 2x，求函数g(x)在区间0，2上的最大值和最小值22 在 ABC 中，A，B，
6、C 所对的边分别为a，b，c，若 A B12，且 a b13，则 cos2B 的值是()A12B.12C32D.3223 在 ABC 中，角 A，B，C 所对的边分别为a，b，c，S表示 ABC 的面积，若acosBbcosAcsinC，S14(b2c2a2)，则 B()A90 B60C45 D301 已知函数 f(x)tan 2x4.(1)求 f(x)的定义域与最小正周期；(2)设 0，4，若 f2 2cos2，求 的大小课标理数 15.C72011 天津卷 【解答】(1)由 2x42k，k Z，得 x8k2，kZ.所以 f(x)的定义域为xR x8k2，kZ.f(x)的最小正周期为2.(2
7、)由 f22cos2，得 tan 42cos2，sin a4cos 42(cos2 sin2)，整理得sin coscos sin2(cos sin)(cos sin)因为 0，4，所以 sin cos 0，因此(cos sin)212，即 sin2 12.由 0，4，得 2 0，2，所以 2 6，即 12.2 在 ABC 中，a，b，c 分别为内角A，B，C 所对的边长，a3，b2，12cos(BC)0，求边 BC 上的高课标文数16.C82011 安徽卷 本题考查两角和的正弦公式，同角三角函数的基本关系，利用正弦定理或余弦定理解三角形，以及三角形的边与角之间的对应大小关系，考查综合运算求解
8、能力【解答】由 12cos(BC)0 和 BC A，得 12cosA0，cosA12，sinA精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页，共 8 页 -文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1
9、F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7
10、 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8
11、I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S
12、7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M
13、1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10
14、H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2学习好资
15、料欢迎下载32.再由正弦定理，得sinBbsinAa22.由 ba 知 BA，所以 B 不是最大角，B2，从而cosB1sin2B22.由上述结果知sinCsin(AB)223212.设边 BC 上的高为h，则有hbsinC312.3 已知 ABC 的一个内角为120，并且三边长构成公差为4 的等差数列，则ABC 的面积为_153【解析】不妨设 A120，cb，则 ab4，cb4，于是 cos120 b2 b42 b422b b412，解得 b10，所以 c6.所以 S12bcsin120 153.4 在 ABC 中，若 b5，B4，tanA2，则 sinA_；a_.【解析】因为 tanA2，
16、所以 sinA2 55；再由正弦定理有：asinAbsinB，即a2 55522，可得 a2 10.5 在 ABC 中，若 b5，B4，sinA13，则 a_.课标文数 9.C82011 北京卷 523【解析】由正弦定理有：asinAbsinB，即a13522，得 a5 23.6 ABC 的内角 A、B、C 的对边分别为a、b、c，asinAcsinC2asinCbsinB.(1)求 B；(2)若 A75，b2，求 a，c.【解答】由正弦定理得a2c22acb2.由余弦定理得b2a2c22accosB.故 cosB22，因此 B45.(2)sinAsin(30 45)sin30 cos45 c
17、os30 sin45 264.故 absinAsinB26213，c bsinCsinB2sin60 sin45 6.课标理数 14.C8图 15 7 如图 15，ABC 中，ABAC2，BC23，点 D 在 BC 边上，ADC45，则AD 的长度等于 _2【解析】在 ABC 中，由余弦定理，有cosCAC2BC2AB22AC BC232222 332，则 ACB30.在ACD 中，由正弦定理，有ADsinCACsinADC，ADAC sin30 sin45 212222，即 AD 的长度等于2.8 若 ABC 的面积为3，BC2，C60，则边 AB 的长度等于 _课标文数 14.C82011
18、 福建卷 2【解析】方法一：由SABC12AC BCsinC，得精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页，共 8 页 -文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10
19、K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 Z
20、B9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9
21、D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文
22、档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU
23、1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A
24、7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2学习好资料欢迎下载12AC 2sin6
25、0 3，解得 AC2.由余弦定理，得AB2AC2BC22AC BCcos60 2222222124，AB2，即边 AB 的长度等于2.方法二：由SAB C12AC BCsinC，得12AC 2sin60 3，解得 AC2.ACBC2,又 ACB60，ABC 是等边三角形，AB2，即边 AB 的长度等于2.9 设 ABC 的内角 A、B、C 所对的边分别为a、b、c，已知 a1，b 2，cosC14.(1)求 ABC 的周长；(2)求 cos(AC)的值课标理数 16.C82011 湖北卷 【解答】(1)c2a2b2 2abcosC144144，c2，ABC 的周长为 abc1225.(2)co
26、sC14，sinC1cos2C1142154，sinAasinCc1542158.ac，A0)，由余弦定理，有cos120 52x27210 x，整理得x25x240，解得 x3，或 x 8(舍去)，即 BC3 所以 SABC12AB BCsinB1253sin120 1253321534.14 在 ABC 中，内角 A，B，C 的对边分别为a，b，c.已知cosA 2cosCcosB2cab.(1)求sinCsinA的值；(2)若 cosB14，ABC 的周长为 5，求 b 的长【解答】(1)由正弦定理，设asinAbsinBcsinCk.则2cab2ksinCksinAksinB2sinC
27、sinAsinB.所以原等式可化为cosA2cosCcosB2sinCsinAsinB.即(cosA2cosC)sinB(2sinCsinA)cosB，化简可得 sin(AB)2sin(BC)，又因为ABC ，所以原等式可化为sinC2sinA，因此sinCsinA2.(2)由正弦定理及sinCsinA2 得 c2a，由余弦定理及cosB14得b2a2c22accosBa24a24a2144a2.所以 b 2a.又 abc 5.从而 a1，因此 b2.15 在 ABC 中，角 A，B，C 所对的边分别为a，b，c.已知 sinAsinCpsinB(pR)，且 ac14b2.(1)当 p54，b
28、 1 时，求 a，c 的值；(2)若角 B 为锐角，求p 的取值范围【解答】(1)由题设并利用正弦定理，得ac54，ac14，解得a 1，c14，或a14，c1.(2)由余弦定理，b2a2c22accosB(ac)22ac2accosBp2b212b212b2cosB，即p23212cosB，因为 0cosB1，得 p232，2，由题设知p 0，所以62p2.16 在 ABC 中，角 A，B，C 的对边分别是a，b，c，已知 3acosAccosBbcosC.(1)求cosA 的值；(2)若 a1，cosBcosC2 33，求边 c 的值【解答】(1)由余弦定理b2a2c22accosB，c2
29、a2b22abcosC，精品资料-欢迎下载-欢迎下载 名师归纳-第 5 页，共 8 页 -文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编
30、码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A
31、1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D
32、7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C
33、8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3
34、S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9
35、M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2学习好资料欢迎下载有 ccosBbcosCa，代入已知条件得3acosAa，即
36、 cosA13.(2)由 cosA13得 sinA223，则 cosB cos(AC)13cosC223sinC，代入 cosBcosC2 33，得 cosC2sinC3，从而得 sin(C)1，其中 sin 33，cos 63，0 2.则 C 2，于是 sinC63，由正弦定理得casinCsinA32.17 在 ABC 中，内角 A，B，C 的对边分别为a，b，c.已知 BC,2b3a.(1)求 cosA 的值；(2)求 cos 2A4的值课标文数 16.C92011 天津卷 【解答】(1)由 BC，2b3a，可得 cb32a.所以 cosAb2c2a22bc34a234a2a2232a3
37、2a13.(2)因为 cosA13，A(0，)，所以 sinA1 cos2A223，故 cos2A2cos2A179.sin2A2sinAcosA4 29.所以 cos 2A4cos2Acos4sin2Asin4 79224292287 218.18 设 aR，f(x)cosx(asinxcosx)cos22x 满足 f 3f(0)求函数 f(x)在4，1124上的最大值和最小值大纲理数 16.C92011 重庆卷 【解答】f(x)asinxcosxcos2xsin2xa2sin2xcos2x.由 f 3f(0)得32a212 1，解得 a2 3.因此 f(x)3sin2xcos2x2sin2
38、x6.当 x4，3时，2x63，2，f(x)为增函数，当x3，1124时,2x62，34，f(x)为减函数所以f(x)在4，1124上的最大值为f32.又因 f43，f11242，故 f(x)在4，1124上的最小值为f11242.19 设函数 f(x)sinxcosx3cos(x)cos x(xR)(1)求 f(x)的最小正周期；(2)若函数 yf(x)的图象按 b4，32平移后得到函数yg(x)的图象，求 yg(x)在 0，4上的最大值 大纲文数 18.C92011 重庆卷 【解答】(1)f(x)12sin2x3cos2x12sin2x32(1cos2x)12sin2x32cos2x32s
39、in 2x332.故 f(x)的最小正周期为T22.(2)依题意 g(x)f x432sin 2 x433232sin 2x63.当 x 0，4时，2x6 6，3，g(x)为增函数，所以 g(x)在 0，4上的最大值为g4332.精品资料-欢迎下载-欢迎下载 名师归纳-第 6 页，共 8 页 -文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K
40、3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB
41、9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D
42、10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档
43、编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1
44、A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7
45、D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5
46、C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2学习好资料欢迎下载20 函数 f(x)2cos2x3sin2x(xR)的最小正周期和最大值分别为()A2，3B2，1C，3D，1 21 函数 f(x)Asin(x )A0，0，|2的部分图象如图所示(1)求 f(x)的最小正周期及解析式；(2)设 g(x)f(x)cos 2x，求函数g(x)在区间0，2上的最大值和最小值22 在 ABC 中，A，B，C 所对的边分别为a，b，c，若 A B12，且 a b13，则 cos2B 的值是()精品资料-欢迎下载-欢迎下载 名
47、师归纳-第 7 页，共 8 页 -文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 Z
48、B9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9
49、D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文
50、档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU1A1F7A7D7 HC5C8I10K3S7 ZB9M1V9D10H2文档编码:CU