高中数学必修第一册 三角恒等变换 单元整体教学设计

高中数学必修第一册三角恒等变换单元整体教学设计一、教材与内容分析【基础·背景】“三角恒等变换”是《普通高中数学课程标准（2017年版2020年修订）》中“函数”主线内容的重要组成部分，隶属于“三角函数”主题。本单元位于人教A版必修第一册第五章“三角函数”的第五节，是在学生系统学习了三角函数的概念、图象与性质以及两角和与差公式的基础上，对三角运算系统的深化与拓展。它既是对三角公式逻辑体系的进一步完善，也是连接三角函数与代数运算的桥梁，更是解决后续课程中解三角形、向量代数、解析几何以及物理中简谐运动、交流电等实际问题的重要工具。【重要·地位】本单元的教学内容主要分为两个核心板块：一是基于两角和与差公式，推导出二倍角公式，并由此引申出半角公式、积化和差、和差化积等公式的探究（虽不要求记忆，但强调推导过程）；二是通过对“辅助角公式”的提炼，实现将形如asin⁡x+bcos⁡xa\sinx+b\cosxasinx+bcosx的表达式合并为一个角的正弦（或余弦）形式，从而统一研究三角函数的图象与性质。这一过程不仅涵盖了公式之间的内在联系与逻辑脉络，更蕴含着丰富的数学思想方法，如转化与化归、换元思想、方程思想、数形结合等。【难点·挑战】与代数恒等变换（如因式分解、多项式运算）不同，三角恒等变换的对象不仅是运算符号和字母，更重要的是“角”与“函数名”。学生在学习过程中面临的主要挑战在于：第一，公式繁多，结构相似，容易混淆，难以形成系统的知识网络；第二，变换的目标意识不强，面对一个表达式不知从何下手，缺乏“看角、看名、看结构”的分析策略；第三，对变换过程中体现的数学思想理解不深，往往陷入盲目套用公式的误区。二、学情分析【基础·认知】授课对象为高中一年级学生。在知识储备上，学生已经掌握了任意角三角函数的定义、同角三角函数基本关系式以及诱导公式，并刚刚学习了两角和与差的正弦、余弦、正切公式，具备了推导新公式的基本工具。在能力层面，学生具备了一定的逻辑推理能力和运算能力，但面对较为复杂的恒等变形时，思维的灵活性和深刻性还有待提升。他们对数学公式的认知多停留在“记忆”层面，对于公式的“来龙去脉”以及公式之间的“血缘关系”缺乏系统性的梳理。【高频考点·心理】从学习心理来看，学生对本章内容普遍存在“又爱又恨”的复杂情感。爱的是公式推导演绎时的逻辑美感，恨的是变换技巧的灵活多变与不易掌握。因此，教学设计的核心不在于让学生死记硬背更多的公式，而在于通过典型问题，引导学生经历公式的再发现、再创造过程，掌握三角变换的基本思想方法，从而减轻记忆负担，提升解决问题的能力。三、教学目标设定基于课程标准与学情分析，确定本单元的教学目标如下：1.【基础】知识与技能：能从两角和与差的正弦、余弦、正切公式出发，推导出二倍角的正弦、余弦、正切公式，并了解其内在联系。能运用二倍角公式进行简单的化简、求值与恒等证明。通过探索，了解半角公式、积化和差、和差化积公式的推导过程（不要求记忆），体会换元与方程思想在其中的应用。掌握辅助角公式asin⁡x+bcos⁡x=a2+b2sin⁡(x+φ)a\sinx+b\cosx=\sqrt{a^2+b^2}\sin(x+\varphi)asinx+bcosx=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​sin(x+φ)（其中tan⁡φ=ba\tan\varphi=\frac{b}{a}tanφ=ab​）的推导过程，并能运用它研究三角函数的周期、最值、单调区间等性质。2.【重要】过程与方法：经历从特殊到一般、从已知到未知的公式发现与推导过程，培养逻辑推理与数学抽象素养。通过对比、分析三角恒等变换与代数恒等变换的异同，提炼“看角、看名、看结构”的变换策略，提升数学运算与直观想象素养。在解决实际问题（如最值问题、化简问题）的过程中，体会转化与化归、数形结合、方程思想等数学思想方法的运用，增强模型意识和应用意识。3.【核心】情感、态度与价值观：在公式的相互推导与联系中，感受数学知识的整体性与系统性，体会数学的逻辑美与严谨美。通过自主探究与合作交流，克服学习困难，树立学好数学的自信心，培养勇于探索、严谨求实的科学精神。四、教学重点与难点【重点】引导学生以十一个基本公式为依据，经历二倍角公式的推导过程，并以此作为基本训练，学习三角变换的内容、思路和方法。掌握辅助角公式的推导及其在解决三角函数综合问题中的应用。【难点】认识三角变换的本质特点——变“角”、变“名”、变“式”，并能自觉地运用数学思想方法（如化归思想、整体代换思想）指导变换过程的设计，提高从整体上把握变换方向的能力。五、教学实施过程（核心环节）本单元教学设计共安排4个课时。教学过程以“问题链”驱动，引导学生在“做中学”、“思中悟”。第一课时：二倍角公式的探索与生成（一）温故知新，引发思考教师通过提问引导学生回顾两角和与差的正弦、余弦、正切公式：sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\betasin(α+β)=sinαcosβ+cosαsinβcos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta\sin\alpha\sin\betacos(α+β)=cosαcosβ−sinαsinβtan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1\tan\alpha\tan\beta}tan(α+β)=1−tanαtanβtanα+tanβ​随后，设置一个特殊化的问题情境：在这些公式中，如果令β=α\beta=\alphaβ=α，公式会变成什么形式？这标志着什么？由此引出本节课的核心问题——二倍角公式的探究。（二）自主探究，公式生成1.【基础·推导】学生独立在练习本上完成对sin⁡2α\sin2\alphasin2α、cos⁡2α\cos2\alphacos2α、tan⁡2α\tan2\alphatan2α的推导。教师巡视，指导学生用准确的数学语言表达推导过程。学生得出：sin⁡2α=2sin⁡αcos⁡α\sin2\alpha=2\sin\alpha\cos\alphasin2α=2sinαcosαcos⁡2α=cos⁡2α−sin⁡2α\cos2\alpha=\cos^2\alpha\sin^2\alphacos2α=cos2α−sin2αtan⁡2α=2tan⁡α1−tan⁡2α\tan2\alpha=\frac{2\tan\alpha}{1\tan^2\alpha}tan2α=1−tan2α2tanα​2.【重要·变式】教师追问：对于cos⁡2α=cos⁡2α−sin⁡2α\cos2\alpha=\cos^2\alpha\sin^2\alphacos2α=cos2α−sin2α，能否利用同角三角函数关系sin⁡2α+cos⁡2α=1\sin^2\alpha+\cos^2\alpha=1sin2α+cos2α=1对它进行变形，用只含cos⁡α\cos\alphacosα或只含sin⁡α\sin\alphasinα的形式表示？学生通过小组讨论，推导出另外两种常用形式：cos⁡2α=2cos⁡2α−1\cos2\alpha=2\cos^2\alpha1cos2α=2cos2α−1cos⁡2α=1−2sin⁡2α\cos2\alpha=12\sin^2\alphacos2α=1−2sin2α教师强调：这三种形式同等重要，在不同情境下各有妙用，尤其是后两种形式，是实现“降幂”或“升幂”变换的关键。（三）公式辨析，深化理解教师通过问题串引导学生深入理解：问题1：二倍角公式中的“二倍”关系是绝对的吗？4α4\alpha4α可以看作是2α2\alpha2α的二倍吗？α\alphaα可以看作是α2\frac{\alpha}{2}2α​的二倍吗？【难点·突破】引导学生认识到“倍”是相对的，蕴含着换元思想。例如：sin⁡4α=2sin⁡2αcos⁡2α\sin4\alpha=2\sin2\alpha\cos2\alphasin4α=2sin2αcos2αcos⁡α=cos⁡2α2−sin⁡2α2=2cos⁡2α2−1=1−2sin⁡2α2\cos\alpha=\cos^2\frac{\alpha}{2}\sin^2\frac{\alpha}{2}=2\cos^2\frac{\alpha}{2}1=12\sin^2\frac{\alpha}{2}cosα=cos22α​−sin22α​=2cos22α​−1=1−2sin22α​这为后续学习半角公式埋下伏笔。问题2：观察公式的结构特征，cos⁡2α\cos2\alphacos2α的三种形式分别有什么功能？师生共同总结：从左到右是“升幂缩角”，从右到左是“降幂扩角”。（四）应用举例，巩固新知例1：【给角求值】求下列各式的值：（1）sin⁡15∘cos⁡15∘\sin15^\circ\cos15^\circsin15∘cos15∘（2）cos⁡2π8−sin⁡2π8\cos^2\frac{\pi}{8}\sin^2\frac{\pi}{8}cos28π​−sin28π​（3）2tan⁡22.5∘1−tan⁡222.5∘\frac{2\tan22.5^\circ}{1\tan^222.5^\circ}1−tan222.5∘2tan22.5∘​例2：【给值求值】已知sin⁡2α=513\sin2\alpha=\frac{5}{13}sin2α=135​，π4<α<π2\frac{\pi}{4}<\alpha<\frac{\pi}{2}4π​<α<2π​，求sin⁡4α\sin4\alphasin4α、cos⁡4α\cos4\alphacos4α、tan⁡4α\tan4\alphatan4α的值。设计意图：例1是公式的直接逆用，强化对公式结构的敏感度。例2需要先判断角的范围，再选择恰当的公式，培养运算的规范性和准确性。第二课时：二倍角公式的变式与应用（一）复习引入，搭建台阶复习上节课的二倍角公式，特别回顾cos⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos2\alpha=2\cos^2\alpha1=12\sin^2\alphacos2α=2cos2α−1=1−2sin2α。教师提出问题：能否将这两个公式变形，用cos⁡2α\cos2\alphacos2α来表示cos⁡2α\cos^2\alphacos2α和sin⁡2α\sin^2\alphasin2α？学生很容易得出：cos⁡2α=1+cos⁡2α2\cos^2\alpha=\frac{1+\cos2\alpha}{2}cos2α=21+cos2α​sin⁡2α=1−cos⁡2α2\sin^2\alpha=\frac{1\cos2\alpha}{2}sin2α=21−cos2α​这就是重要的“降幂公式”。它实现了二次幂向一次角的转化，是简化三角函数解析式、研究周期与最值的利器。（二）半角公式的探源教师引导学生进行换元：令α=θ2\alpha=\frac{\theta}{2}α=2θ​，则上面的降幂公式可以写成：cos⁡2θ2=1+cos⁡θ2\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}cos22θ​=21+cosθ​sin⁡2θ2=1−cos⁡θ2\sin^2\frac{\theta}{2}=\frac{1\cos\theta}{2}sin22θ​=21−cosθ​进而得到：cos⁡θ2=±1+cos⁡θ2\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}cos2θ​=±21+cosθ​<pathd="M98390l00c4,6.7,10,10,18,10Hv40H1013.1s83.4,268,264.1,840c180.7,572,277,876.3,289,913c4.7,4.7,12.7,7,24,7s12,0,12,0c1.3,3.3,3.7,11.7,7,25c35.3,125.3,106.7,373.3,214,744c10,12,21,25,33,39s32,39,32,39c6,5.3,15,14,27,26s25,30,25,30c26.7,32.7,52,63,76,91s52,60,52,60s208,722,208,722c56,175.3,126.3,397.3,211,666c84.7,268.7,153.8,488.2,207.5,658.5c53.7,170.3,84.5,266.8,92.5,289.5zMhv40hz">​sin⁡θ2=±1−cos⁡θ2\sin\frac{\theta}{2}=\pm\sqrt{\frac{1\cos\theta}{2}}sin2θ​=±21−cosθ​<pathd="M98390l00c4,6.7,10,10,18,10Hv40H1013.1s83.4,268,264.1,840c180.7,572,277,876.3,289,913c4.7,4.7,12.7,7,24,7s12,0,12,0c1.3,3.3,3.7,11.7,7,25c35.3,125.3,106.7,373.3,214,744c10,12,21,25,33,39s32,39,32,39c6,5.3,15,14,27,26s25,30,25,30c26.7,32.7,52,63,76,91s52,60,52,60s208,722,208,722c56,175.3,126.3,397.3,211,666c84.7,268.7,153.8,488.2,207.5,658.5c53.7,170.3,84.5,266.8,92.5,289.5zMhv40hz">​tan⁡θ2=±1−cos⁡θ1+cos⁡θ\tan\frac{\theta}{2}=\pm\sqrt{\frac{1\cos\theta}{1+\cos\theta}}tan2θ​=±1+cosθ1−cosθ​<pathd="M98390l00c4,6.7,10,10,18,10Hv40H1013.1s83.4,268,264.1,840c180.7,572,277,876.3,289,913c4.7,4.7,12.7,7,24,7s12,0,12,0c1.3,3.3,3.7,11.7,7,25c35.3,125.3,106.7,373.3,214,744c10,12,21,25,33,39s32,39,32,39c6,5.3,15,14,27,26s25,30,25,30c26.7,32.7,52,63,76,91s52,60,52,60s208,722,208,722c56,175.3,126.3,397.3,211,666c84.7,268.7,153.8,488.2,207.5,658.5c53.7,170.3,84.5,266.8,92.5,289.5zMhv40hz">​【难点·警示】教师强调：这里的符号由θ2\frac{\theta}{2}2θ​所在的象限决定，不能遗漏。虽然课程标准对半角公式的记忆不作要求，但这一推导过程本身就是极好的思维训练，它清晰地展示了“倍半相对”和“换元思想”的力量。（三）积化和差与和差化积的思维体操教师展示一个高观点下的问题：如何将两个三角函数的乘积形式转化为和差形式？引导学生从两角和与差公式入手：sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\betasin(α+β)=sinαcosβ+cosαsinβsin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha\beta)=\sin\alpha\cos\beta\cos\alpha\sin\betasin(α−β)=sinαcosβ−cosαsinβ将两式相加，得：sin⁡(α+β)+sin⁡(α−β)=2sin⁡αcos⁡β\sin(\alpha+\beta)+\sin(\alpha\beta)=2\sin\alpha\cos\betasin(α+β)+sin(α−β)=2sinαcosβ所以：sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]\sin\alpha\cos\beta=\frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha\beta)]sinαcosβ=21​[sin(α+β)+sin(α−β)]同理，通过两式相减或对其他公式进行组合，可以推导出积化和差的另外三个公式。接着，通过换元，令x=α+βx=\alpha+\betax=α+β，y=α−βy=\alpha\betay=α−β，则可以反解出α=x+y2\alpha=\frac{x+y}{2}α=2x+y​，β=x−y2\beta=\frac{xy}{2}β=2x−y​，代入积化和差公式，即可得到和差化积公式。【核心·思想】教师总结：这一系列的推导，让我们看到了数学公式之间严密的逻辑体系，以及“加减法消元”与“换元构造”这两种处理复杂代数关系的普适方法。我们不需要记忆最终公式，但必须掌握这种“执果索因”的探究过程。（四）综合训练例3：化简sin⁡50∘(1+3tan⁡10∘)\sin50^\circ(1+\sqrt{3}\tan10^\circ)sin50∘(1+3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​tan10∘)。设计意图：引导学生观察角与角之间的关系（50°与10°），尝试将正切化为正弦与余弦，引入辅助角或使用和差化积思想，综合运用所学知识。第三课时：辅助角公式——化繁为简的利器（一）情境创设，问题驱动播放一段简谐运动的视频或展示物理中交流电的波形图。提出问题：如何研究一个形如y=3sin⁡x+4cos⁡xy=3\sinx+4\cosxy=3sinx+4cosx的函数的图象与性质？学生尝试：这个函数既含有正弦，又含有余弦，无法直接从图象变换的角度理解其周期和最值。教师引导：面对一个陌生的、复杂的表达式，我们的基本策略是将其转化为熟悉的、简单的形式。我们最熟悉的形式是什么？是y=Asin⁡(ωx+φ)y=A\sin(\omegax+\varphi)y=Asin(ωx+φ)或y=Acos⁡(ωx+φ)y=A\cos(\omegax+\varphi)y=Acos(ωx+φ)。那么，能否将asin⁡x+bcos⁡xa\sinx+b\cosxasinx+bcosx合并成一个角的一个三角函数呢？（二）合作探究，公式推导1.【思路启发】我们曾经遇到过类似的问题吗？比如，sin⁡x+cos⁡x\sinx+\cosxsinx+cosx可以怎么处理？引导学生回忆起特殊形式：sin⁡x+cos⁡x=2(22sin⁡x+22cos⁡x)=2sin⁡(x+π4)\sinx+\cosx=\sqrt{2}(\frac{\sqrt{2}}{2}\sinx+\frac{\sqrt{2}}{2}\cosx)=\sqrt{2}\sin(x+\frac{\pi}{4})sinx+cosx=2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​(22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​sinx+22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​cosx)=2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​sin(x+4π​)2.【类比迁移】对于一般的asin⁡x+bcos⁡xa\sinx+b\cosxasinx+bcosx，能否也构造出两角和的正弦公式形式？分析：我们需要找到一个角φ\varphiφ，使得cos⁡φ=aa2+b2\cos\varphi=\frac{a}{\sqrt{a^2+b^2}}cosφ=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​a​，sin⁡φ=ba2+b2\sin\varphi=\frac{b}{\sqrt{a^2+b^2}}sinφ=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​b​。这样，asin⁡x+bcos⁡x=a2+b2(aa2+b2sin⁡x+ba2+b2cos⁡x)a\sinx+b\cosx=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sinx+\frac{b}{\sqrt{a^2+b^2}}\cosx)asinx+bcosx=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​(a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​a​sinx+a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​b​cosx)=a2+b2(sin⁡xcos⁡φ+cos⁡xsin⁡φ)=\sqrt{a^2+b^2}(\sinx\cos\varphi+\cosx\sin\varphi)=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​(sinxcosφ+cosxsinφ)=a2+b2sin⁡(x+φ)=\sqrt{a^2+b^2}\sin(x+\varphi)=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​sin(x+φ)其中，φ\varphiφ所在的象限由点(a,b)(a,b)(a,b)的坐标决定，且tan⁡φ=ba\tan\varphi=\frac{b}{a}tanφ=ab​。3.【重要·拓展】教师提问：是否也可以合并成余弦形式？引导学生得出：asin⁡x+bcos⁡x=a2+b2cos⁡(x−θ)a\sinx+b\cosx=\sqrt{a^2+b^2}\cos(x\theta)asinx+bcosx=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​cos(x−θ)，其中cos⁡θ=ba2+b2\cos\theta=\frac{b}{\sqrt{a^2+b^2}}cosθ=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​b​，sin⁡θ=aa2+b2\sin\theta=\frac{a}{\sqrt{a^2+b^2}}sinθ=a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​a​，tan⁡θ=ab\tan\theta=\frac{a}{b}tanθ=ba​。（三）公式应用，能力提升例4：【高频考点】已知函数f(x)=3sin⁡2x−cos⁡2xf(x)=\sqrt{3}\sin2x\cos2xf(x)=3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​sin2x−cos2x。（1）求f(x)f(x)f(x)的最小正周期和最大值；（2）求f(x)f(x)f(x)的单调递增区间；（3）求f(x)f(x)f(x)在区间[0,π2][0,\frac{\pi}{2}][0,2π​]上的值域。处理过程：第一步：化简。引导学生确定a=3a=\sqrt{3}a=3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​，b=−1b=1b=−1，则a2+b2=2\sqrt{a^2+b^2}=2a2+b2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=2。设cos⁡φ=32\cos\varphi=\frac{\sqrt{3}}{2}cosφ=23<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，sin⁡φ=−12\sin\varphi=\frac{1}{2}sinφ=−21​，可知φ\varphiφ是第四象限角，取φ=−π6\varphi=\frac{\pi}{6}φ=−6π​。所以f(x)=2sin⁡(2x−π6)f(x)=2\sin(2x\frac{\pi}{6})f(x)=2sin(2x−6π​)。第二步：求解。将2x−π62x\frac{\pi}{6}2x−6π​视为一个整体，利用正弦函数的图象与性质进行求解。教师规范板书，强调整体代换思想。例5：【难点·应用】求函数y=sin⁡x−1cos⁡x−2y=\frac{\sinx1}{\cosx2}y=cosx−2sinx−1​的值域。设计意图：此题具有挑战性