高一数学必修第二册“三角恒等变换”章末复习教学设计

高一数学必修第二册“三角恒等变换”章末复习教学设计一、教学分析（一）教材分析【基础】三角恒等变换是高中数学必修课程的核心内容之一，承载着发展学生逻辑推理、数学运算和数学抽象素养的重要功能。苏教版高中数学必修第二册将本章安排在平面向量之后、解三角形之前，具有承上启下的作用。一方面，它借助向量数量积推导两角差的余弦公式，体现了向量作为工具的价值；另一方面，恒等变换的结果又为后续学习解三角形、三角函数图象与性质以及物理中的简谐运动奠定坚实基础。本章的核心是一系列三角恒等变换公式，包括两角和与差的正弦、余弦、正切公式，二倍角公式，以及由此推导出的半角公式、积化和差与和差化积公式（视学情选学）。这些公式的内在逻辑紧密，通过变换可以实现不同形式间的转化，是培养学生逻辑推理和数学运算能力的优质载体。本章节内容在历年高考中占有重要地位，通常以选择题、填空题或解答题第一问的形式出现，分值约为5—12分，【高频考点】主要集中在公式的正用、逆用、变形用以及恒等变换在化简、求值、证明中的应用。（二）学情分析授课对象为高一学生，他们已经系统学习了任意角三角函数的概念、同角三角函数基本关系式、诱导公式以及两角和与差的基本公式，具备了一定的运算能力和推理经验。但学生在学习过程中容易出现以下问题：公式繁多，记忆混淆，无法灵活选择公式；对公式的结构特征把握不准，在恒等变形中方向性不强；综合运用多个公式时，运算准确率不高；面对复杂问题，缺乏转化与化归的意识。此外，学生虽然初步接触了向量法推导公式，但跨学科联系尚显薄弱。因此，章末复习课需要在唤醒旧知、构建网络的基础上，通过典型例题引导学生梳理解题思路，提炼通性通法，并在综合应用中提升数学核心素养。（三）教学目标1.知识与技能：熟练掌握两角和与差、二倍角的正弦、余弦、正切公式，理解公式的推导过程及内在联系；能运用公式进行三角函数的化简、求值和恒等式证明；了解辅助角公式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)的推导与应用，【重要】能将其转化为单一三角函数形式解决最值、周期等问题。2.过程与方法：通过自主梳理、合作探究，构建三角恒等变换公式网络，体会转化与化归、数形结合的思想；在典例剖析中，掌握公式的正用、逆用、变形用，学会观察角、名、结构的差异，选择恰当的变换策略。3.情感、态度与价值观：感受数学公式的对称美与简洁美，体验恒等变换在解决实际问题（如物理中的合成振动）中的价值，增强应用意识；在克服运算困难的过程中，培养严谨求实的科学态度和锲而不舍的钻研精神。（四）教学重难点【重点】熟练掌握三角恒等变换公式，并能灵活运用于化简、求值与证明。【难点】公式的灵活选择与变形应用，尤其是辅助角公式的运用以及恒等变换在综合问题中的转化策略。二、教学策略与方法本节课采用“问题驱动—自主建构—典例导学—变式提升”的教学模式。课前布置学生用思维导图梳理本章公式，课上通过小组交流完善知识网络。教学中以问题链引导学生深入思考，精选典型例题，层层递进，注重一题多解与多题归一，揭示恒等变换的本质。同时，借助多媒体展示公式推导的几何背景，如向量旋转、单位圆上的三角函数线，渗透数形结合思想。对于易错点，通过错例辨析强化认知。此外，引入物理中的简谐振动合成问题，体现跨学科视野，激发学习兴趣。三、教学过程设计本单元复习共安排3课时。第1课时：知识梳理与公式网络构建，基础训练；第2课时：典型例题剖析与变式拓展，能力提升；第3课时：章末检测与试卷讲评，反馈矫正。（一）第1课时：知识梳理，构建网络（45分钟）1.课前准备：布置学生自主回顾本章内容，用思维导图的形式整理所有公式，并思考公式间的推导逻辑。同时完成预学单上的几个简单求值题，唤醒记忆。2.课堂实施（1）导入（5分钟）教师展示一组物理问题：两个简谐振动y1=3sin⁡(2t)y_1=3\sin(2t)y1​=3sin(2t)，y2=4cos⁡(2t)y_2=4\cos(2t)y2​=4cos(2t)，它们的合振动y=y1+y2y=y_1+y_2y=y1​+y2​能否用一个正弦函数表示？如何求出合振动的振幅和初相？引发学生思考，点明三角恒等变换在跨学科中的应用价值，引出复习主题。（2）公式网络共建（15分钟）小组内交流课前绘制的思维导图，互相补充完善。教师选取若干份有代表性的作品投影展示，引导学生评价其逻辑性和完整性。在此基础上，师生共同提炼出公式体系：①两角和与差公式（基础）：cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\betacos(α−β)=cosαcosβ+sinαsinβcos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta\sin\alpha\sin\betacos(α+β)=cosαcosβ−sinαsinβsin⁡(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−cosαsinβtan⁡(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​tan⁡(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​②二倍角公式（重要）：sin⁡2α=2sin⁡αcos⁡α\sin2\alpha=2\sin\alpha\cos\alphasin2α=2sinαcosαcos⁡2α=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos2\alpha=\cos^2\alpha\sin^2\alpha=2\cos^2\alpha1=12\sin^2\alphacos2α=cos2α−sin2α=2cos2α−1=1−2sin2αtan⁡2α=2tan⁡α1−tan⁡2α\tan2\alpha=\frac{2\tan\alpha}{1\tan^2\alpha}tan2α=1−tan2α2tanα​③由二倍角公式变形得到的降幂公式（常用）：sin⁡2α=1−cos⁡2α2\sin^2\alpha=\frac{1\cos2\alpha}{2}sin2α=21−cos2α​，cos⁡2α=1+cos⁡2α2\cos^2\alpha=\frac{1+\cos2\alpha}{2}cos2α=21+cos2α​④辅助角公式（重点）：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​（a>0a>0a>0），或=a2+b2cos⁡(x−φ)=\sqrt{a^2+b^2}\cos(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">​cos(x−φ)形式。⑤半角公式（选学，视班级情况补充）：sin⁡α2=±1−cos⁡α2\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1\cos\alpha}{2}}sin2α​=±21−cosα​<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">​，cos⁡α2=±1+cos⁡α2\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}cos2α​=±21+cosα​<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">​，符号由α2\frac{\alpha}{2}2α​所在象限决定。⑥积化和差与和差化积公式（供学有余力学生了解）：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)]等。教师强调公式间的逻辑链条：以两角差的余弦公式为源头，通过换元推导出两角和与差的其他公式，再通过令β=α\beta=\alphaβ=α得到二倍角公式，进而衍生出降幂公式、半角公式等。并指出公式的记忆方法：如余弦公式“同名相乘，符号相反”；正弦公式“异名相乘，符号相同”；正切公式的分子分母结构等。（3）基础训练（20分钟）设计一组直接运用公式的题目，帮助学生巩固记忆，规范书写。①求值：sin⁡75∘\sin75^\circsin75∘，cos⁡15∘\cos15^\circcos15∘，tan⁡105∘\tan105^\circtan105∘。②已知cos⁡α=35\cos\alpha=\frac{3}{5}cosα=53​，α∈(0,π2)\alpha\in(0,\frac{\pi}{2})α∈(0,2π​)，求sin⁡(α+π3)\sin(\alpha+\frac{\pi}{3})sin(α+3π​)。③化简：sin⁡20∘cos⁡10∘+cos⁡20∘sin⁡10∘cos⁡20∘cos⁡10∘−sin⁡20∘sin⁡10∘\frac{\sin20^\circ\cos10^\circ+\cos20^\circ\sin10^\circ}{\cos20^\circ\cos10^\circ\sin20^\circ\sin10^\circ}cos20∘cos10∘−sin20∘sin10∘sin20∘cos10∘+cos20∘sin10∘​。④已知tan⁡α=2\tan\alpha=2tanα=2，求tan⁡2α\tan2\alphatan2α和tan⁡(α−π4)\tan(\alpha\frac{\pi}{4})tan(α−4π​)。学生独立完成，教师巡视指导，对典型错误进行板演纠正。例如第④题中，学生易将tan⁡2α\tan2\alphatan2α公式记错，教师强调公式结构，并引导学生利用tan⁡2α=2tan⁡α1−tan⁡2α\tan2\alpha=\frac{2\tan\alpha}{1\tan^2\alpha}tan2α=1−tan2α2tanα​求解。（4）课堂小结与作业（5分钟）教师总结：今天我们从公式的源头出发，梳理了三角恒等变换的公式体系，并通过基础训练初步运用。请同学们课后熟记公式，完成课后练习题，并思考：这些公式在化简求值中如何灵活选用？布置作业：课本章末复习题A组1—5题；整理错题本。（二）第2课时：典例剖析，深化理解（45分钟）1.导入（3分钟）回顾上节课的公式网络，点明公式的正用、逆用和变形用是解题的关键。展示一道高考题：已知sin⁡α+cos⁡α=15\sin\alpha+\cos\alpha=\frac{1}{5}sinα+cosα=51​，0<α<π0<\alpha<\pi0<α<π，求tan⁡α\tan\alphatanα的值。引导学生思考：如何利用恒等变换求解？自然过渡到本节课。2.典例剖析（1）例1：给值求值问题已知cos⁡(α−π6)+sin⁡α=435\cos(\alpha\frac{\pi}{6})+\sin\alpha=\frac{4\sqrt{3}}{5}cos(α−6π​)+sinα=543<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">​​，求sin⁡(α+7π6)\sin(\alpha+\frac{7\pi}{6})sin(α+67π​)的值。【分析】本题观察角的关系：α+7π6=(α−π6)+π\alpha+\frac{7\pi}{6}=(\alpha\frac{\pi}{6})+\piα+67π​=(α−6π​)+π，因此只需求出cos⁡(α−π6)\cos(\alpha\frac{\pi}{6})cos(α−6π​)或sin⁡(α−π6)\sin(\alpha\frac{\pi}{6})sin(α−6π​)即可。但已知条件中有cos⁡(α−π6)\cos(\alpha\frac{\pi}{6})cos(α−6π​)和sin⁡α\sin\alphasinα，需将sin⁡α\sin\alphasinα用α−π6\alpha\frac{\pi}{6}α−6π​表示：sin⁡α=sin⁡[(α−π6)+π6]\sin\alpha=\sin[(\alpha\frac{\pi}{6})+\frac{\pi}{6}]sinα=sin[(α−6π​)+6π​]，展开后与cos⁡(α−π6)\cos(\alpha\frac{\pi}{6})cos(α−6π​)合并，可求出sin⁡(α−π6)\sin(\alpha\frac{\pi}{6})sin(α−6π​)或cos⁡(α−π6)\cos(\alpha\frac{\pi}{6})cos(α−6π​)。教师引导学生尝试不同的思路，并比较优劣。最终规范解答：解：由已知得cos⁡(α−π6)+sin⁡α=cos⁡(α−π6)+sin⁡[(α−π6)+π6]\cos(\alpha\frac{\pi}{6})+\sin\alpha=\cos(\alpha\frac{\pi}{6})+\sin[(\alpha\frac{\pi}{6})+\frac{\pi}{6}]cos(α−6π​)+sinα=cos(α−6π​)+sin[(α−6π​)+6π​]=cos⁡(α−π6)+sin⁡(α−π6)cos⁡π6+cos⁡(α−π6)sin⁡π6=\cos(\alpha\frac{\pi}{6})+\sin(\alpha\frac{\pi}{6})\cos\frac{\pi}{6}+\cos(\alpha\frac{\pi}{6})\sin\frac{\pi}{6}=cos(α−6π​)+sin(α−6π​)cos6π​+cos(α−6π​)sin6π​=cos⁡(α−π6)+32sin⁡(α−π6)+12cos⁡(α−π6)=\cos(\alpha\frac{\pi}{6})+\frac{\sqrt{3}}{2}\sin(\alpha\frac{\pi}{6})+\frac{1}{2}\cos(\alpha\frac{\pi}{6})=cos(α−6π​)+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(α−6π​)+21​cos(α−6π​)=32cos⁡(α−π6)+32sin⁡(α−π6)=\frac{3}{2}\cos(\alpha\frac{\pi}{6})+\frac{\sqrt{3}}{2}\sin(\alpha\frac{\pi}{6})=23​cos(α−6π​)+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(α−6π​)。令r=(32)2+(32)2=94+34=3r=\sqrt{(\frac{3}{2})^2+(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{3}r=(23​)2+(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">​​)2<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">​=49​+43​<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">​=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">​，则上式可化为3sin⁡(α−π6+φ)\sqrt{3}\sin(\alpha\frac{\pi}{6}+\varphi)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">​sin(α−6π​+φ)，其中tan⁡φ=3\tan\varphi=\sqrt{3}tanφ=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">​，即φ=π3\varphi=\frac{\pi}{3}φ=3π​。所以3sin⁡(α−π6+π3)=435\sqrt{3}\sin(\alpha\frac{\pi}{6}+\frac{\pi}{3})=\frac{4\sqrt{3}}{5}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">​sin(α−6π​+3π​)=543<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">​​，即sin⁡(α+π6)=45\sin(\alpha+\frac{\pi}{6})=\frac{4}{5}sin(α+6π​)=54​。从而sin⁡(α+7π6)=sin⁡(α+π6+π)=−sin⁡(α+π6)=−45\sin(\alpha+\frac{7\pi}{6})=\sin(\alpha+\frac{\pi}{6}+\pi)=\sin(\alpha+\frac{\pi}{6})=\frac{4}{5}sin(α+67π​)=sin(α+6π​+π)=−sin(α+6π​)=−54​。【重要】本例展示了角的变换与辅助角公式的综合应用，提醒学生注意整体思想。（2）例2：化简问题化简：2cos⁡2α−12tan⁡(π4−α)sin⁡2(π4+α)\frac{2\cos^2\alpha1}{2\tan(\frac{\pi}{4}\alpha)\sin^2(\frac{\pi}{4}+\alpha)}2tan(4π​−α)sin2(4π​+α)2cos2α−1​。【分析】观察式子结构：分子2cos⁡2α−1=cos⁡2α2\cos^2\alpha1=\cos2\alpha2cos2α−1=cos2α；分母中有tan⁡(π4−α)\tan(\frac{\pi}{4}\alpha)tan(4π​−α)和sin⁡2(π4+α)\sin^2(\frac{\pi}{4}+\alpha)sin2(4π​+α)，注意到π4−α\frac{\pi}{4}\alpha4π​−α与π4+α\frac{\pi}{4}+\alpha4π​+α互余，可考虑利用诱导公式转化。尝试将tan⁡(π4−α)=sin⁡(π4−α)cos⁡(π4−α)\tan(\frac{\pi}{4}\alpha)=\frac{\sin(\frac{\pi}{4}\alpha)}{\cos(\frac{\pi}{4}\alpha)}tan(4π​−α)=cos(4π​−α)sin(4π​−α)​，且sin⁡(π4−α)=cos⁡(π4+α)\sin(\frac{\pi}{4}\alpha)=\cos(\frac{\pi}{4}+\alpha)sin(4π​−α)=cos(4π​+α)，cos⁡(π4−α)=sin⁡(π4+α)\cos(\frac{\pi}{4}\alpha)=\sin(\frac{\pi}{4}+\alpha)cos(4π​−α)=sin(4π​+α)，则tan⁡(π4−α)=cos⁡(π4+α)sin⁡(π4+α)\tan(\frac{\pi}{4}\alpha)=\frac{\cos(\frac{\pi}{4}+\alpha)}{\sin(\frac{\pi}{4}+\alpha)}tan(4π​−α)=sin(4π​+α)cos(4π​+α)​。代入分母，化简可得结果。学生尝试化简，教师提示关键步骤：解：原式=cos⁡2α2⋅cos⁡(π4+α)sin⁡(π4+α)⋅sin⁡2(π4+α)=cos⁡2α2cos⁡(π4+α)sin⁡(π4+α)=\frac{\cos2\alpha}{2\cdot\frac{\cos(\frac{\pi}{4}+\alpha)}{\sin(\frac{\pi}{4}+\alpha)}\cdot\sin^2(\frac{\pi}{4}+\alpha)}=\frac{\cos2\alpha}{2\cos(\frac{\pi}{4}+\alpha)\sin(\frac{\pi}{4}+\alpha)}=2⋅sin(4π​+α)cos(4π​+α)​⋅sin2(4π​+α)cos2α​=2cos(4π​+α)sin(4π​+α)cos2α​。分母2cos⁡(π4+α)sin⁡(π4+α)=sin⁡(2(π4+α))=sin⁡(π2+2α)=cos⁡2α2\cos(\frac{\pi}{4}+\alpha)\sin(\frac{\pi}{4}+\alpha)=\sin(2(\frac{\pi}{4}+\alpha))=\sin(\frac{\pi}{2}+2\alpha)=\cos2\alpha2cos(4π​+α)sin(4π​+α)=sin(2(4π​+α))=sin(2π​+2α)=cos2α。所以原式=cos⁡2αcos⁡2α=1=\frac{\cos2\alpha}{\cos2\alpha}=1=cos2αcos2α​=1。【基础】本例训练了公式的逆用（二倍角公式逆用）和互余角的转化，化简过程中要敏锐发现结构特点。（3）例3：恒等式证明求证：1+sin⁡2θ−cos⁡2θ1+sin⁡2θ+cos⁡2θ=tan⁡θ\frac{1+\sin2\theta\cos2\theta}{1+\sin2\theta+\cos2\theta}=\tan\theta1+sin2θ+cos2θ1+sin2θ−cos2θ​=tanθ。【分析】从左向右证明，思路一：从左边出发，利用二倍角公式将sin⁡2θ\sin2\thetasin2θ、cos⁡2θ\cos2\thetacos2θ用单角表示，然后化简。思路二：将cos⁡2θ\cos2\thetacos2θ用1−2sin⁡2θ12\sin^2\theta1−2sin2θ或2cos⁡2θ−12\cos^2\theta12cos2θ−1代入，看能否约分。选择较简捷的方法。证法一：左边=1+2sin⁡θcos⁡θ−(1−2sin⁡2θ)1+2sin⁡θcos⁡θ+(2cos⁡2θ−1)=2sin⁡θcos⁡θ+2sin⁡2θ2sin⁡θcos⁡θ+2cos⁡2θ=\frac{1+2\sin\theta\cos\theta(12\sin^2\theta)}{1+2\sin\theta\cos\theta+(2\cos^2\theta1)}=\frac{2\sin\theta\cos\theta+2\sin^2\theta}{2\sin\theta\cos\theta+2\cos^2\theta}=1+2sinθcosθ+(2cos2θ−1)1+2sinθcosθ−(1−2sin2θ)​=2sinθcosθ+2cos2θ2sinθcosθ+2sin2θ​=2sin⁡θ(cos⁡θ+sin⁡θ)2cos⁡θ(sin⁡θ+cos⁡θ)=sin⁡θcos⁡θ=tan⁡θ=\frac{2\sin\theta(\cos\theta+\sin\theta)}{2\cos\theta(\sin\theta+\cos\theta)}=\frac{\sin\theta}{\cos\theta}=\tan\theta=2cosθ(sinθ+cosθ)2sinθ(cosθ+sinθ)​=cosθsinθ​=tanθ=右边。证法二：左边=1−cos⁡2θ+sin⁡2θ1+cos⁡2θ+sin⁡2θ=2sin⁡2θ+2sin⁡θcos⁡θ2cos⁡2θ+2sin⁡θcos⁡θ=2sin⁡θ(sin⁡θ+cos⁡θ)2cos⁡θ(cos⁡θ+sin⁡θ)=tan⁡θ=\frac{1\cos2\theta+\sin2\theta}{1+\cos2\theta+\sin2\theta}=\frac{2\sin^2\theta+2\sin\theta\cos\theta}{2\cos^2\theta+2\sin\theta\cos\theta}=\frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)}=\tan\theta=1+cos2θ+sin2θ1−cos2θ+sin2θ​=2cos2θ+2sinθcosθ2sin2θ+2sinθcosθ​=2cosθ(cosθ+sinθ)2sinθ(sinθ+cosθ)​=tanθ。【热点】恒等式证明的关键在于观察等式两边的差异，选择恰当的公式统一函数名或角。3.变式拓展（15分钟）（1）变式1：已知sin⁡α+cos⁡α=15\sin\alpha+\cos\alpha=\frac{1}{5}sinα+cosα=51​，0<α<π0<\alpha<\pi0<α<π，求tan⁡α\tan\alphatanα的值。引导学生利用sin⁡α+cos⁡α=2sin⁡(α+π4)\sin\alpha+\cos\alpha=\sqrt{2}\sin(\alpha+\frac{\pi}{4})sinα+cosα=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(α+4π​)先求sin⁡(α+π4)\sin(\alpha+\frac{\pi}{4})sin(α+4π​)，再结合范围求α\alphaα，进而得tan⁡α\tan\alphatanα；或者两边平方，结合sin⁡2α+cos⁡2α=1\sin^2\alpha+\cos^2\alpha=1sin2α+cos2α=1求出2sin⁡αcos⁡α=−24252\sin\alpha\cos\alpha=\frac{24}{25}2sinαcosα=−2524​，进而构造tan⁡α+cot⁡α\tan\alpha+\cot\alphatanα+cotα或利用sin⁡α−cos⁡α\sin\alpha\cos\alphasinα−cosα求解。通过多种解法加深对恒等变换的理解。（2）变式2：已知函数f(x)=sin⁡2x+3sin⁡xcos⁡xf(x)=\sin^2x+\sqrt{3}\sinx\cosxf(x)=sin2x+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">​sinxcosx，求f(x)f(x)f(x)的周期、单调区间及最大值。先引导学生利用降幂公式化简：f(x)=1−cos⁡2x2+32sin⁡2x=12+32sin⁡2x−12cos⁡2xf(x)=\frac{1\cos2x}{2}+\frac{\sqrt{3}}{2}\sin2x=\frac{1}{2}+\frac{\sqrt{3}}{2}\sin2x\frac{1}{2}\cos2xf(x)=21−cos2x​+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">​​sin2x=21​+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">​​sin2x−21​cos2x，再用辅助角公式合并为f(x)=12+sin⁡(2x−π6)f(x)=\frac{1}{2}+\sin(2x\frac{\pi}{6})f(x)=21​+sin(2x−6π​)。然后讨论周期T=πT=\piT=π，单调区间及最值。本题是三角恒等变换与三角函数性质的综合应用，【重要】体现变换的工具性。4.课堂小结与作业（2分钟）总结本节课的核心思想：观察差异（角、名、结构），选择公式，灵活变换。强调化归思想与整体代换。布置作业：完成章末复习题B组1—3题；预习下一节内容。（三）第3课时：章末检测，反馈评价（45分钟）1.检测环节（30分钟）发放预先设计的章末检测卷（满分50分，时间30分钟），题目覆盖本章核心知识点，难度适中，注重基础与能力并重。检测卷结构如下：【基础题】（每题5分，共20分）（1）sin⁡15∘cos⁡15∘\sin15^\circ\cos15^\circsin15∘cos15∘的值为______。（2）已知cos⁡α=45\cos\alpha=\frac{4}{5}cosα=54​，α∈(−π2,0)\alpha\in(\frac{\pi}{2},0)α∈(−2π​,0)，则sin⁡(α+π3)\sin(\alpha+\frac{\pi}{3})sin(α+3π​)=。（3）函数y=sin⁡2xcos⁡2xy=\sin2x\cos2xy=sin2xcos2x的最小正周期是。（4）化简tan⁡10∘+tan⁡20∘1−tan⁡10∘tan⁡20∘\frac{\tan10^\circ+\tan20^\circ}{1\tan10^\circ\tan20^\circ}1−tan10∘tan20∘tan10∘+tan20∘​的结果为______。【中档题】（每题6分，共18分）（5）已知sin⁡(α−π4)=7210\sin(\alpha\frac{\pi}{4})=\frac{7\sqrt{2}}{10}sin(α−4π​)=1072<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⁡2α=725\cos2\alpha=\frac{7}{25}cos2α=257​，求sin⁡α\sin\alphasinα的值。（6）求函数f(x)=cos⁡2x+sin⁡xf(x)=\cos2x+\sinxf(x)=cos2x+sinx的最大值和最小值。（7）求证：sin⁡2α1+cos⁡2α⋅cos⁡α1+cos⁡α=tan⁡α2\frac{\sin2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}1+cos2αsin2α​⋅1+cosαcosα​=tan2α​。【拓展题】（12分）（8）如图，在平面直角坐标系中，以xxx轴正半轴为始边作两个锐角α,β\alpha,\betaα,β，它们的终边分别与单位圆交于A,BA,BA,B两点，已知A,BA,BA,B的横坐标分别为1010,255\frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5}1010<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">​​,525<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">​​。①求tan⁡(α+β)\tan(\alpha+\beta)tan(α+β)的值；②求α+2β\alpha+2\betaα+2β的值。学生独立完成，教师巡视，观察学生答题情况，记录典型错误。2.讲评环节（15分钟）针对检测中暴露的问题，重点讲解第5、7、8题。（1）第5题：已知sin⁡(α−π4)=7210\sin(\alpha\frac{\pi}{4})=\frac{7\sqrt{2}}{10}sin(α−4π​)=1072<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⁡2α=725\cos2\alpha=\frac{7}{25}cos2α=257​，求sin⁡α\sin\alphasinα。分析思路：可先利用cos⁡2α=1−2sin⁡2α\cos2\alpha=12\sin^2\alphacos2α=1−2sin2α或2cos⁡2α−12\cos^2\alpha12cos2α−1求出sin⁡α\sin\alphasinα或cos⁡α\cos\alphacosα的值，但要注意符号的确定。同时也可将sin⁡(α−π4)\sin(\alpha\frac{\pi}{4})sin(α−4π​)展开，与cos⁡2α\cos2\alphacos2α联立求解。展示两种解法，强调角的范围对符号的影响。（2）第7题：证明恒等式。学生可能从左边化到右边有困难，教师引导：左边有sin⁡2α1+cos⁡2α\frac{\sin2\alpha}{1+\cos2\alpha}1+cos2αsin2α​，可先化简为tan⁡α\tan\alphatanα，则左边变为tan⁡α⋅cos⁡α1+cos⁡α\tan\alpha\cdot\frac{\cos\alpha}{1+\cos\alpha}tanα⋅1+cosαcosα​，再化为sin⁡α1+cos⁡α\frac{\sin\alpha}{1+\cos\alpha}1+cosαsinα​，最后利用半角公式tan⁡α2=sin⁡α1+cos⁡α\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}tan2α​=1+cosαsinα​得证。强调半角公式的灵活应用。（3）第8题：结合三角函数定义与恒等变换。先由定义求出cos⁡α=1010\cos\alpha=\frac{\sqrt{10}}{10}cosα=1010<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⁡β=255\cos\beta=\frac{2\sqrt{5}}{5}cosβ=525<