《第10讲 三角函数的恒等变形与求值.pptVIP

三角函数的恒等变形与求值 一、学习目标 1、了解用向量的数量积推导出两角差的余弦公式的过程. 2、能利用已知条件,正确合理地运用三角恒等变形公式进行三角函数式的化简、求值及恒等式证明. 二、课前诊断 三、例题探究 三、例题探究 新华中学 蒋同山 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2．已知 ， 则 = 问题：角是否已确定？除明显条件外，有无隐含条件？如何将条件简化？ 点评：三个基本形式 三者之间的转化要很熟练. 注意角的范围可避免讨论. -1 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 变题1：若 为第二象限角， ，则 = 问题：从条件和结论的角度看，化简方向是否已确定？有无函数名的差异？如何转化？ Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 变题2： 问题：从条件和结论的角度看，二倍角的正弦往单角的正弦转化还需要什么中间环节？有无符号问题？ Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 点评：用两角和的正切公式与韦达定理 化简求值. -3 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 变题：若 ， 则 = 点评：本题需求正弦值,显然必须切化弦; 然后利用正余弦平方和为1将式子简化;最后逆用二倍角公式即得. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 点评：利用三角恒等变换化为一个角的三角函数从而由三角函数的有界性得值域. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 点评：先用倍角公式,再用同角三角函数的基本关系式,最后用和角三角函数的公式. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 三角函数式化简目标 1.项数尽可能少; 2.三角函数名称尽可能少; 3.角尽可能小和少; 4.次数尽可能低; 5.分母尽可能不含三角式; 6.尽可能不带根号; 7.能求出值的求出值. 小结： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.