高一数学两角与与差的三角函数余树宝.ppt

P(x,y) r x=r ·cosα y=r ·sinα P(r ·cosα , r ·sinα) O x y 我们把y/r、x/r、y/x、x/y、r/x、r/y定义为α的六个三 角函数 特别地，r=1时，点P的坐标为（cos α,sin α) C:\sketch Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. 第三章 两角和与差的三角函数， 解斜三角形更多资源 一、两角和与差的三角函数 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. 不查表，求cos（ –435°） 的值. 解：cos(–435 ° ) =cos435 ° =cos(360 ° +75 °)=cos75 ° Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. §3.1 两角和与差的三角函数 1.两角和与差的余弦 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. cos(α+β)=cosαcosβ–sinαsinβ 在平面直角坐标系xOy内,作单位圆,并作 α 、 β 和–β角,使α角的始边为Ox，交圆O于P1, 终边交圆O于P2;β角的始边为OP2,终边交圆O于 P3; – β角的始边为OP1,终边交圆O于P4; 此时,P1.P2.P3.P4的坐标分别为P1(1,0) , P2(cosα,sinα), P3(cos(α+β),sin(α+β) ), P4(cos(–β), sin(–β)). 由︱P1P3 ︱= ︱P2P4︱及两点间距离公式, 得： [cos(α+β)–1]2+sin2(α+β)=[cos(–β)–cosα]2+[sin(–β)–sinα] 2. 整理得: cos(α+β)=cosαcosβ–sinαsinβ. 证明:如图所示 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. cos(α+β)=cosαcosβ–sinαsinβ 公式的结构特征: 左边是复角α+β 的余弦,右边是单角α、β的余弦积与正弦积的差. cos(α–β)= cosαcosβ+sinαsinβ 公式的结构特征: 左边是复角α+β的余弦,右边是单角α、β 的余弦积 与正弦积的和. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. 例1.不查表,求cos(–435°)的值. 解:cos(– 435 °)=cos75 ° =cos(45 ° +30 °) =cos45 ° ?cos30 ° –sin45 ° ?sin30 ° 应用举例 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd. 不查表,求cos105 °和cos15 °的值. 练习 Evaluation only. Created with Asp