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1、第五节 和、差、倍角的三角函数(1)Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.基础梳理1. 两角和与差的正弦、余弦和正切公式 C(a-b)：cos(a-b)=_； C(a+b)：co
2、s(a+b)=_； S(a+b)：sin(a+b)=_； S(a-b)：sin(a-b)=_；T(a+b)：tan(a+b)=_；T(a-b)：tan(a-b)=_.cos acos b+sin asin b cos acos b-sin asin b sin acos b+cos asin b sin acos b-cos asin b2. 二倍角的正弦、余弦、正切公式 S2a：sin 2a=_； C2a：cos 2a=_=_=_；T2a：tan 2a=_.2sin acos a cos2a-sin 2a2cos2a-1 1-2sin2a Evaluation only.Evaluation
3、 only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.基础达标 1.(2010福建改编)sin43cos13-sin13cos43 2.的值为_解析：原式=sin(43-13)=sin30= 2. sin105cos105的值为_ 解析：sin1
4、05cos105=-sin75cos75=- sin150=- 3. (必修4P97例6改编)若则 解析：因为 所以 因此 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.4. 函数y=s
5、inx+cos 的最大值和最小值分别为_ 解析： 当x=2k +时，ymax=(kZ)当x=2k -时，ymax=5. (必修4P97第7题改编)设 则sin 的值为_解析： 又 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-
6、2011 Aspose Pty Ltd.经典例题 题型一 化简求值【例1】 求2sin50+sin10(1+ tan 10) 的值 分析：50、10、80都不是特殊角，但注意到它们的 和60、90都是特殊角，因此可考虑用和角公式求其 值；另外含有正切函数，切化弦后出现分式，可通过约分 以去掉非特殊角解：原式= sin50cos10+sin10cos(60-10) sin(50+10)= Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created w
7、ith Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty LEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-201
8、1 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.变式1-1 求sin50(1+ tan10)的值. 解析：原式=sin50 题型二 给值求角【例2】 已知 为锐角，向量a=(cos ，sin )，b=(cos ，sin )，c= 求角 的值.若a b=a c=及a，b，c的坐标，可求由a b=a c=分析 出关于a、b的三角函数值，进而求出角解：a b= Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0
9、.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.a c= 由得 由得 又 、 为锐角， 从而 变式2-1 已知 并且 均为锐角，求 解析： 且 、 均为锐角， 又 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created w
10、ith Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.题型三 给值求值【例3】 已知 求sin 的值 分析：对已知条件因式分解，求出tan ，再求得sin 、 cos ，进而求解 解： 又 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created wit
11、h Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.变式3-1(2010 全国)已知 是第二象限的角， 则 解析：由 得 又 得 又 是第二象限的角，所以 链接高考(2010 北京)已知函数f(x)=2cos2x+sin2x-4cosx.(1)求f 的值；(2)求f(x)的最大值和最小值知识准备：1. 知道特殊角的函数值； 2. 会用倍角公式； 3. 会用配方法求函数最值Evaluation only.
12、Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.解析：(1) (2) 因为cosx-1,1， 所以，当cosx=-1时，f(x)取最大值6；当cosx= 时，f(x)取最小值- Evaluation only.Evalu
13、ation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty LEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.
