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1、学习好资料欢迎下载三角函数与平面向量的综合应用1 三角恒等变换(1)公式：同角三角函数基本关系式、诱导公式、和差公式(2)公式应用：注意公式的正用、逆用、变形使用的技巧，观察三角函数式中角之间的联系，式子之间以及式子和公式间的联系(3)注意公式应用的条件、三角函数的符号、角的范围2 三角函数的性质(1)研究三角函数的性质，一般要化为yAsin(x )的形式，其特征：一角、一次、一函数(2)在讨论 yAsin(x )的图象和性质时，要重视两种思想的应用：整体思想和数形结合思想，一般地，可设t x ，yAsin t，通过研究这两个函数的图象、性质达到目的3 解三角形解三角形问题主要有两种题型：一是
2、与三角函数结合起来考查，通过三角变换化简，然后运用正、余弦定理求值；二是与平面向量结合(主要是数量积)，判断三角形形状或结合正、余弦定理求值试题一般为中档题，客观题、解答题均有可能出现4 平面向量平面向量的线性运算，为证明两线平行提供了重要方法平面向量数量积的运算解决了两向量的夹角、垂直等问题特别是平面向量的坐标运算与三角函数的有机结合，体现了向量应用的广泛性1 已知角 终边上一点P(4,3)，则cos2sin cos112sin92的值为 _答案34解析cos2sin cos112sin92sin sin sin cos tan .学习好资料欢迎下载根据三角函数的定义得tan yx34.所以
3、cos2sin cos112 sin9234.2 已知 f(x)sin(x)3cos(x)的一条对称轴为y 轴，且 (0，)，则 _.答案6解析f(x)sin(x)3cos(x)2sin x 3，由 3k 2(kZ)及 (0，)，可得 6.3.如图所示的是函数f(x)Asin(x )B(A0，0，|0，2)图象的一部分，则f(x)的解析式为 _答案f(x)2sin23x61 解析由于最大值和最小值之差等于4，故 A2，B1.由于 2 2sin 1，且|0，2，得 6.由图象知()2k 2(kZ)，得 2k23(k Z)又22，0 1.23.函数 f(x)的解析式是f(x)2sin23x61.4
4、(2012 四川改编)如图，正方形ABCD 的边长为1，延长 BA 至 E，使 AE1，连接 EC、ED，则 sinCED _.答案1010解析方法一应用两角差的正弦公式求解由题意知，在RtADE 中，AED45，在 RtBCE 中，BE2，BC 1，CE5，则 sin CEB15，cosCEB25.文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1
5、I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8
6、 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7
7、P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N
8、6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:
9、CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U
10、6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10
11、 HT7S1I6K1S8 ZQ3K7P7N10N6文档编码:CK3O6U6K7W10 HT7S1I6K1S8 ZQ3K7P7N10N6学习好资料欢迎下载而 CED45 CEB，sinCEDsin(45 CEB)22(cosCEBsinCEB)2225151010.方法二利用余弦定理及同角三角函数基本关系式求解由题意得ED2，EC12225.在 EDC 中，由余弦定理得cosCEDCE2DE2 DC22CE DE31010，又 0CED，sinCED1cos2CED13101021010.5.如图，在梯形ABCD 中，ADBC，ADAB，AD1，BC2，AB 3，P 是 BC 上的一个动点，当P
12、D PA取得最小值时，tanDPA 的值为 _答案1235解析如图，以A 为原点，建立平面直角坐标系xAy，则 A(0,0)，B(3，0)，C(3,2)，D(0,1)，设 CPD，BPA，P(3，y)(0y2)PD(3,1y)，PA(3，y)，PD PAy2y9 y122354，当 y12时，PD PA取得最小值，此时P 3，12，易知|DP|AP|，.在 ABP 中，tan 3126，文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O
13、9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:
14、CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O
15、9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:
16、CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O
17、9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:
18、CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O
19、9R1 ZB5F9A9C6B4学习好资料欢迎下载tanDPA tan()2tan tan2 11235.题型一三角恒等变换例 1设3 34，sin 435，求sin cos 2 1tan 的值思维启迪：可以先将所求式子化简，寻求和已知条件的联系解方法一由3 34，得12 40.由于3 34，故3 0,0 2，yf(x)的部分图象如图所示，P、Q 分别为该图象的最高点和最低点，点 P 的坐标为(1，A)(1)求 f(x)的最小正周期及的值；(2)若点 R 的坐标为(1,0)，PRQ23，求 A 的值思维启迪：三角函数图象的确定，可以利用图象的周期性、最值、已知点的坐标列方程来解决解(1)由题意得
20、T236.因为 P(1，A)在 yAsin(3x )的图象上，所以 sin(3)1.又因为 0 0，所以A3.探究提高本题确定的值时，一定要考虑的范围；在三角形中利用余弦定理求A 是本题的难点已知函数f(x)Asin x Bcos x(A，B，是常数，0)的最小正周期为2，并且当 x13时，f(x)max2.(1)求 f(x)的解析式；(2)在闭区间214，234上是否存在f(x)的对称轴？如果存在，求出其对称轴方程；如果不存在，请说明理由解(1)因为 f(x)A2B2sin(x )，由它的最小正周期为2，知22，又因为当 x13时，f(x)max 2，知13 2k 2(kZ)，2k 6(k
21、Z)，所以 f(x)2sin x2k 62sin x6.故 f(x)的解析式为f(x)2sin x6.(2)当垂直于 x 轴的直线过正弦曲线的最高点或最低点时，该直线就是正弦曲线的对称轴，令 x6k 2(kZ)，解得 xk13，由214k13234，解得5912k6512，又 kZ，知 k5，由此可知在闭区间214，234上存在 f(x)的对称轴，其方程为x163.题型三三角函数、平面向量、解三角形的综合应用例 3已知向量m3sin x4，1，n cos x4，cos2x4.(1)若 m n1，求 cos23 x 的值；(2)记 f(x)m n，在 ABC 中，角 A，B，C 的对边分别是a，
22、b，c，且满足(2ac)cos Bbcos C，求函数f(A)的取值范围思维启迪：(1)由向量数量积的运算转化成三角函数式，化简求值(2)在ABC 中，求出文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3
23、R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档
24、编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3
25、R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档
26、编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3
27、R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档
28、编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4学习好资料欢迎下载A 的范围，再求f(A)的取值范围解(1)m n3sin x4 cos x4cos2x432sin x21cos x22sinx2612，m n1，sinx2612.cos x312si
29、n2x2612，cos23 x cos x312.(2)(2a c)cos Bbcos C，由正弦定理得(2sin A sin C)cos Bsin Bcos C，2sin Acos Bsin Ccos Bsin Bcos C.2sin Acos Bsin(BC)ABC，sin(B C)sin A0.cos B12，0B，B3.0A23.6A262，sinA2612，1.又 f(x)sinx2612.f(A)sinA2612.故函数 f(A)的取值范围是1，32.探究提高(1)向量是一种解决问题的工具，是一个载体，通常是用向量的数量积运算或性质转化成三角函数问题(2)三角形中的三角函数要结合正
30、弦定理、余弦定理进行转化，注意角的范围对变形过程的影响在 ABC 中，角 A，B，C 的对边分别为a，b，c，且 lg alg blg cos Blg cos A0.(1)判断 ABC 的形状；(2)设向量 m(2a，b)，n(a，3b)，且 mn，(mn)(nm)14，求 a，b，c 的值文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9
31、A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5
32、U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9
33、A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5
34、U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9
35、A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5
36、U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4学习好资料欢迎下载解(1)因为 lg alg blg cos Blg
37、 cos A0，所以abcos Bcos A1，所以 sin 2Asin 2B 且 ab.因为 A，B(0，)且 AB，所以 2A 2B，即 AB2且 AB.所以 ABC 是非等腰的直角三角形(2)由 mn，得 m n0.所以 2a23b20.由(mn)(n m)14，得 n2m214，所以 a29b24a2b214，即 3a28b214.联立 ，解得 a6，b 2.所以 ca2b210.故所求的a，b，c 的值分别为6，2，10.高考中的平面向量、三角函数客观题典例 1：(5 分)(2012 山东)函数 y2sin x63(0 x 9)的最大值与最小值之和为()A23 B0 C 1 D 13
38、 考点分析本题考查三角函数的性质，考查整体思想和数形结合思想解题策略根据整体思想，找出角6x3的范围，再根据图象求函数的最值解析由题意3 x6376.画出 y 2sin x 的图象如图，知，当6x33时，ymin3.当6x32时，ymax2.故 ymaxymin 23.答案A 解后反思(1)函数 y Asin(x )可看作由函数yAsin t 和 tx 构成的复合函数文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A
39、9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U
40、2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A
41、9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U
42、2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A
43、9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U
44、2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A
45、9C6B4学习好资料欢迎下载(2)复合函数的值域即为外层函数的值域，可以通过图象观察得到典例 2：(5 分)(2012 天津)在 ABC 中，A90，AB1，AC2.设点 P，Q 满足 AP AB，AQ(1)AC，R.若 BQ CP 2，则 等于()A.13B.23C.43D2 考点分析本题考查向量的线性运算，考查向量的数量积和运算求解能力解题策略根据平面向量基本定理，将题中的向量BQ，CP分别用向量 AB，AC表示出来，再进行数量积计算解析BQAQAB(1)ACAB，CPAPAC ABAC，BQ CP(1)AC2 AB24(1)3 4 2，即 23.答案B 解后反思(1)利用平面向量基本定理
46、结合向量的线性运算表示向量是向量问题求解的基础；(2)本题在求解过程中利用了方程思想方法与技巧1研究三角函数的图象、性质一定要化成yAsin(x )B 的形式，然后利用数形结合思想求解2三角函数与向量的综合问题，一般情况下向量知识作为一个载体，可以先通过计算转化为三角函数问题再进行求解失误与防范1三角函数式的变换要熟练公式，注意角的范围2向量计算时要注意向量夹角的大小，不要混同于直线的夹角或三角形的内角文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2
47、HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C
48、6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2
49、HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C
50、6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2 HJ3X3R7O9R1 ZB5F9A9C6B4文档编码:CG9I8V2D5U2