高中数学《两角差的余弦公式》教学设计

高中数学《两角差的余弦公式》教学设计一、教材与学情分析（一）教材地位与作用【基础】本节课选自人教A版普通高中课程标准实验教科书数学必修4第三章第一节。两角差的余弦公式是本章最先推导和学习的公式，是整个三角恒等变换的基石。后续的两角和与差的正弦、正切公式，以及二倍角公式，都可以由此公式为基础通过诱导公式或代换推导得出。它沟通了任意角与特殊角、任意角与任意角之间的三角函数关系，是计算三角函数值、化简三角函数式、证明三角恒等式的重要工具，具有承上启下的核心地位。【核心】（二）学情分析【重要】授课对象为高中二年级学生。在此之前，学生已经掌握了任意角三角函数的定义、同角三角函数的基本关系式以及诱导公式，具备了一定的逻辑推理能力和运算求解能力。然而，学生对“恒等变换”的理解还停留在代数运算层面，对“三角变换”中蕴含的“变角、变名、变结构”的思想方法尚不熟悉。在心理层面，学生容易对复杂的公式推导产生畏难情绪，因此教学设计需注重创设问题情境，引导学生主动参与探究过程，从特殊到一般，从具体到抽象，逐步建构知识，化解难点。【难点】二、教学目标设计（一）知识与技能目标【基础】学生能借助单位圆中的三角函数线或向量的数量积等工具，自主推导出两角差的余弦公式cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\betacos(α−β)=cosαcosβ+sinαsinβ。理解公式的结构特征，掌握公式的正用、逆用和变形用。能够运用公式解决简单的三角函数求值、化简问题。（二）过程与方法目标【重要】通过公式的发现与证明过程，让学生体会数形结合、化归与转化的数学思想。经历观察、猜想、验证、证明等数学活动，提升学生的逻辑推理能力和数学抽象素养。在公式的推导过程中，鼓励学生从不同角度思考问题（如三角函数线法、向量法），培养发散思维和批判性思维。（三）情感、态度与价值观目标【核心】通过公式的简洁美与对称美，感受数学的内在魅力。在探究活动中，培养学生勇于探索、敢于质疑的科学精神以及严谨求实的治学态度。通过小组合作与交流，增强团队协作意识。三、教学重难点（一）教学重点【核心】【高频考点】两角差的余弦公式的推导与理解，以及公式的初步应用。（二）教学难点【难点】探究公式的推导方法，尤其是如何构造单位圆上的点，利用两点间距离公式或向量数量积完成证明，并理解其中蕴含的数学思想。四、教学方法与准备（一）教学方法采用“问题链”引导下的探究式教学法，结合启发式讲授与小组合作学习。以问题为驱动，层层递进，引导学生自主建构新知。利用多媒体课件（如几何画板）动态演示，增强直观性，突破难点。（二）教学准备教师准备：多媒体课件（含几何画板动态演示）、导学案。学生准备：复习三角函数的定义、诱导公式、向量数量积等相关知识，预习教材。五、教学实施过程（一）创设情境，引入新知【基础】教师首先提出问题：我们已经能够熟练计算cos⁡60∘\cos60^\circcos60∘、cos⁡45∘\cos45^\circcos45∘等特殊角的三角函数值。那么，如何计算cos⁡15∘\cos15^\circcos15∘的值呢？请同学们思考。学生可能会联想到15∘=60∘−45∘15^\circ=60^\circ45^\circ15∘=60∘−45∘，于是问题转化为求cos⁡(60∘−45∘)\cos(60^\circ45^\circ)cos(60∘−45∘)。教师追问：cos⁡(60∘−45∘)\cos(60^\circ45^\circ)cos(60∘−45∘)是否等于cos⁡60∘−cos⁡45∘\cos60^\circ\cos45^\circcos60∘−cos45∘？引导学生用计算器验证cos⁡15∘≈0.9659\cos15^\circ\approx0.9659cos15∘≈0.9659，而cos⁡60∘−cos⁡45∘=0.5−0.7071=−0.2071\cos60^\circ\cos45^\circ=0.50.7071=0.2071cos60∘−cos45∘=0.5−0.7071=−0.2071，两者显然不相等。此环节通过认知冲突，激发学生的求知欲，自然地引出本节课的核心课题：如何用α\alphaα和β\betaβ的三角函数来表示cos⁡(α−β)\cos(\alpha\beta)cos(α−β)。（二）探究新知，建构公式【核心】本环节是课堂教学的重中之重，将引导学生从几何和代数两个角度进行探究。1.探究路径一：利用单位圆上的三角函数线（几何法）教师引导：三角函数与单位圆紧密相连。我们能否在单位圆中构造出角α\alphaα、β\betaβ以及α−β\alpha\betaα−β，并利用几何关系（如两点间距离）来寻找等量关系？（1）构造图形：在平面直角坐标系xOyxOyxOy中作出单位圆OOO。以xxx轴非负半轴为始边，分别作角α\alphaα和β\betaβ（假设α>β>0\alpha>\beta>0α>β>0），设它们的终边与单位圆的交点分别为点P1P_1P1​和点P2P_2P2​。则P1P_1P1​的坐标为(cos⁡α,sin⁡α)(\cos\alpha,\sin\alpha)(cosα,sinα)，P2P_2P2​的坐标为(cos⁡β,sin⁡β)(\cos\beta,\sin\beta)(cosβ,sinβ)。再作角α−β\alpha\betaα−β，设其终边与单位圆的交点为AAA，则AAA的坐标为(cos⁡(α−β),sin⁡(α−β))(\cos(\alpha\beta),\sin(\alpha\beta))(cos(α−β),sin(α−β))。同时，也可以将角α−β\alpha\betaα−β理解为从射线OP2OP_2OP2​旋转到射线OP1OP_1OP1​所形成的角。（2）寻找等量关系：连接P1P2P_1P_2P1​P2​和OAOAOA。容易发现，在单位圆中，弧P1P2P_1P_2P1​P2​所对的圆心角为∣α−β∣|\alpha\beta|∣α−β∣，而弧A(1,0)A(1,0)A(1,0)所对的圆心角也为∣α−β∣|\alpha\beta|∣α−β∣。因此，根据圆心角相等，则对应的弦长相等，可得弦P1P2P_1P_2P1​P2​的长度等于弦A(1,0)A(1,0)A(1,0)的长度。（3）坐标运算：根据两点间距离公式：∣P1P2∣2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2|P_1P_2|^2=(\cos\alpha\cos\beta)^2+(\sin\alpha\sin\beta)^2∣P1​P2​∣2=(cosα−cosβ)2+(sinα−sinβ)2=cos⁡2α−2cos⁡αcos⁡β+cos⁡2β+sin⁡2α−2sin⁡αsin⁡β+sin⁡2β=\cos^2\alpha2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha2\sin\alpha\sin\beta+\sin^2\beta=cos2α−2cosαcosβ+cos2β+sin2α−2sinαsinβ+sin2β=(cos⁡2α+sin⁡2α)+(cos⁡2β+sin⁡2β)−2(cos⁡αcos⁡β+sin⁡αsin⁡β)=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=(cos2α+sin2α)+(cos2β+sin2β)−2(cosαcosβ+sinαsinβ)=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)=22(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=2−2(cosαcosβ+sinαsinβ)点AAA的坐标为(cos⁡(α−β),sin⁡(α−β))(\cos(\alpha\beta),\sin(\alpha\beta))(cos(α−β),sin(α−β))，点(1,0)(1,0)(1,0)的坐标为(1,0)(1,0)(1,0)。则：∣OA∣2=[cos⁡(α−β)−1]2+[sin⁡(α−β)−0]2|OA|^2=[\cos(\alpha\beta)1]^2+[\sin(\alpha\beta)0]^2∣OA∣2=[cos(α−β)−1]2+[sin(α−β)−0]2=cos⁡2(α−β)−2cos⁡(α−β)+1+sin⁡2(α−β)=\cos^2(\alpha\beta)2\cos(\alpha\beta)+1+\sin^2(\alpha\beta)=cos2(α−β)−2cos(α−β)+1+sin2(α−β)=[cos⁡2(α−β)+sin⁡2(α−β)]+1−2cos⁡(α−β)=[\cos^2(\alpha\beta)+\sin^2(\alpha\beta)]+12\cos(\alpha\beta)=[cos2(α−β)+sin2(α−β)]+1−2cos(α−β)=2−2cos⁡(α−β)=22\cos(\alpha\beta)=2−2cos(α−β)（4）建立等式：由∣P1P2∣2=∣OA∣2|P_1P_2|^2=|OA|^2∣P1​P2​∣2=∣OA∣2得：2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)=2−2cos⁡(α−β)22(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=22\cos(\alpha\beta)2−2(cosαcosβ+sinαsinβ)=2−2cos(α−β)化简即得：cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\betacos(α−β)=cosαcosβ+sinαsinβ此方法直观地展示了公式的几何背景，体现了数形结合思想。教师需强调公式对任意角α,β\alpha,\betaα,β都成立，因为上述推导中α,β\alpha,\betaα,β并未受特殊范围限制。2.探究路径二：利用向量的数量积（代数法）【重要】教师启发：我们刚刚学习了向量，向量的数量积与角度有着天然的联系。能否用向量工具来推导这个公式呢？（1）构造向量：在单位圆上，设以xxx轴非负半轴为始边，角α\alphaα和β\betaβ的终边与单位圆的交点分别为P1(cos⁡α,sin⁡α)P_1(\cos\alpha,\sin\alpha)P1​(cosα,sinα)和P2(cos⁡β,sin⁡β)P_2(\cos\beta,\sin\beta)P2​(cosβ,sinβ)。那么，向量OP1→=(cos⁡α,sin⁡α)\overrightarrow{OP_1}=(\cos\alpha,\sin\alpha)OP1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​=(cosα,sinα)，向量OP2→=(cos⁡β,sin⁡β)\overrightarrow{OP_2}=(\cos\beta,\sin\beta)OP2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​=(cosβ,sinβ)。这两个向量的夹角就是∣α−β∣|\alpha\beta|∣α−β∣（或2π−∣α−β∣2\pi|\alpha\beta|2π−∣α−β∣，但余弦值相同）。（2）数量积公式：根据向量数量积的定义，有：OP1→⋅OP2→=∣OP1→∣⋅∣OP2→∣⋅cos⁡⟨OP1→,OP2→⟩=1×1×cos⁡(α−β)\overrightarrow{OP_1}\cdot\overrightarrow{OP_2}=|\overrightarrow{OP_1}|\cdot|\overrightarrow{OP_2}|\cdot\cos\langle\overrightarrow{OP_1},\overrightarrow{OP_2}\rangle=1\times1\times\cos(\alpha\beta)OP1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​⋅OP2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​=∣OP1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​∣⋅∣OP2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​∣⋅cos⟨OP1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​,OP2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​⟩=1×1×cos(α−β)（3）坐标运算：根据向量数量积的坐标表示，有：OP1→⋅OP2→=cos⁡αcos⁡β+sin⁡αsin⁡β\overrightarrow{OP_1}\cdot\overrightarrow{OP_2}=\cos\alpha\cos\beta+\sin\alpha\sin\betaOP1​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​⋅OP2​<pathd="M0241v40hc47.335.3847811012816.73227.763..3.22.7.54.31.3.52.3.5307.36.71120.2.815.52.52.31.74.25.55.511.5213.35.727114114.744.73984..5s73.760..5c6295.7911s39911c45.315.38540..5s58.374..5c4.7148.327..36.73.210.85.512.52.31.77.52.515.52..7211102210..783.367151.zm00v40hv40z">​=cosαcosβ+sinαsinβ（4）得出结论：由以上两式相等，即得：cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\betacos(α−β)=cosαcosβ+sinαsinβ向量法的推导更为简洁、严谨，充分体现了向量这一工具的强大功能，也为后续学习空间向量和解析几何打下基础。教师应引导学生对比两种方法的优劣，体会数学知识的内部联系。【核心】（三）深化理解，记忆公式【基础】引导学生观察公式的结构特征：左边是两角差的余弦，右边是这两个角的同名三角函数乘积之和，即“余余正正，符号相反”（这里的符号相反是指公式中是加号，而左边是差角）。为了加深记忆，可引导学生将其与完全平方公式等进行类比，但更要强调其独特的结构。教师通过几何画板动态演示，验证公式对任意角（如α\alphaα为钝角，β\betaβ为负角）均成立，进一步巩固学生对公式普适性的认识。（四）例题精讲，巩固新知【高频考点】本环节通过典型例题，引导学生掌握公式的正用、逆用和变形用。1.公式的正用：给值求值例1：利用两角差的余弦公式，求cos⁡15∘\cos15^\circcos15∘及cos⁡75∘\cos75^\circcos75∘的值。解：cos⁡15∘=cos⁡(45∘−30∘)=cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘\cos15^\circ=\cos(45^\circ30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circcos15∘=cos(45∘−30∘)=cos45∘cos30∘+sin45∘sin30∘=22×32+22×12=6+24=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}=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">​​×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">​​+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">​​×21​=46<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="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⁡75∘=cos⁡(45∘+30∘)=cos⁡45∘cos⁡30∘−sin⁡45∘sin⁡30∘=6−24\cos75^\circ=\cos(45^\circ+30^\circ)=\cos45^\circ\cos30^\circ\sin45^\circ\sin30^\circ=\frac{\sqrt{6}\sqrt{2}}{4}cos75∘=cos(45∘+30∘)=cos45∘cos30∘−sin45∘sin30∘=46<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="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">​​（此处将75∘75^\circ75∘视为45∘45^\circ45∘与30∘30^\circ30∘的和，为下节课学习两角和的余弦公式埋下伏笔。）练习：求cos⁡105∘\cos105^\circcos105∘的值。引导学生将105∘105^\circ105∘拆分成两个特殊角的差或和。2.公式的逆用：化简求值例2：化简cos⁡40∘cos⁡20∘+sin⁡40∘sin⁡20∘\cos40^\circ\cos20^\circ+\sin40^\circ\sin20^\circcos40∘cos20∘+sin40∘sin20∘。分析：引导学生观察式子的结构，发现其恰好符合公式右边“余余正正相加”的形式，因此可以逆用公式。解：原式=cos⁡(40∘−20∘)=cos⁡20∘=\cos(40^\circ20^\circ)=\cos20^\circ=cos(40∘−20∘)=cos20∘。例3：求cos⁡80∘cos⁡35∘+cos⁡10∘cos⁡55∘\cos80^\circ\cos35^\circ+\cos10^\circ\cos55^\circcos80∘cos35∘+cos10∘cos55∘的值。分析：此题不能直接逆用，需要先利用诱导公式进行转化。cos⁡10∘=sin⁡80∘\cos10^\circ=\sin80^\circcos10∘=sin80∘，cos⁡55∘=sin⁡35∘\cos55^\circ=\sin35^\circcos55∘=sin35∘。解：原式=cos⁡80∘cos⁡35∘+sin⁡80∘sin⁡35∘=cos⁡(80∘−35∘)=cos⁡45∘=22=\cos80^\circ\cos35^\circ+\sin80^\circ\sin35^\circ=\cos(80^\circ35^\circ)=\cos45^\circ=\frac{\sqrt{2}}{2}=cos80∘cos35∘+sin80∘sin35∘=cos(80∘−35∘)=cos45∘=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">​​。【重要】3.公式的变式应用：给值求角例4：已知sin⁡α=45\sin\alpha=\frac{4}{5}sinα=54​，α∈(π2,π)\alpha\in(\frac{\pi}{2},\pi)α∈(2π​,π)，cos⁡β=−513\cos\beta=\frac{5}{13}cosβ=−135​，β\betaβ是第三象限角，求cos⁡(α−β)\cos(\alpha\beta)cos(α−β)的值。分析：这是典型的“给值求值”问题，且涉及象限符号判断，是高考的常见题型。【高频考点】解：由sin⁡α=45\sin\alpha=\frac{4}{5}sinα=54​，α∈(π2,π)\alpha\in(\frac{\pi}{2},\pi)α∈(2π​,π)，得cos⁡α=−1−sin⁡2α=−1−(45)2=−35\cos\alpha=\sqrt{1\sin^2\alpha}=\sqrt{1(\frac{4}{5})^2}=\frac{3}{5}cosα=−1−sin2α<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMhv40hz">​=−1−(54​)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">​=−53​。由cos⁡β=−513\cos\beta=\frac{5}{13}cosβ=−135​，β\betaβ是第三象限角，得sin⁡β=−1−cos⁡2β=−1−(−513)2=−1213\sin\beta=\sqrt{1\cos^2\beta}=\sqrt{1(\frac{5}{13})^2}=\frac{12}{13}sinβ=−1−cos2β<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMhv40hz">​=−1−(−135​)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">​=−1312​。cos⁡(α−β)=cos⁡