# 高中数学/函数与三角/两角和与差的三角函数公式

## 基础知识

### 和角公式与差角公式

#### 公式推导

${\displaystyle P_{1}=(1,0),P_{2}=(\cos(a+b),\sin(a+b)),P_{3}=(\cos a,\sin a),P_{4}=(\cos(-b),\sin(-b))}$

${\displaystyle {\begin{array}{l}|P_{1}P_{2}|^{2}=(\cos(a+b)-1)^{2}+\sin ^{2}(a+b)\\|P_{3}P_{4}|^{2}=(\cos(-b)-\cos a)^{2}+(\sin(-b)-\sin a)^{2}\end{array}}}$

${\displaystyle {\begin{array}{l}(\cos(a+b)-1)^{2}+\sin ^{2}(a+b)=(\cos(-b)-\cos a)^{2}+(\sin(-b)-\sin a)^{2}\\\Rightarrow \cos ^{2}(a+b)-2\cos(a+b)+1+\sin ^{2}(a+b)=\cos ^{2}(-b)-2\cos(-b)\cos a+\cos ^{2}a+\sin ^{2}(-b)-2\sin(-b)\sin a+\sin ^{2}a\\\Rightarrow (\cos ^{2}(a+b)+\sin ^{2}(a+b))-2\cos(a+b)+1=\cos ^{2}b-2\cos b\cos a+\cos ^{2}a+\sin ^{2}b+2\sin b\sin a+\sin ^{2}a\\\Rightarrow 1-2\cos(a+b)+1=(\sin ^{2}b+\cos ^{2}b)+(\sin ^{2}a+\cos ^{2}a)-2\cos b\cos a+2\sin b\sin a\\\Rightarrow 2-2\cos(a+b)=1+1-2\cos b\cos a+2\sin b\sin a\\\Rightarrow -2\cos(a+b)=-2\cos b\cos a+2\sin b\sin a\\\Rightarrow \cos(a+b)=\cos a\cos b-\sin a\sin b\end{array}}}$

${\displaystyle \cos(a-b)=\cos a\cos(-b)-\sin a\sin(-b)=\cos a\cos b+\sin a\sin b}$

${\displaystyle \cos({\frac {\pi }{2}}-a)=\cos {\frac {\pi }{2}}\cos a+\sin {\frac {\pi }{2}}\sin a=0\cdot \cos a+1\cdot \sin a=\sin a}$
${\displaystyle \cos a=\sin({\frac {\pi }{2}}-a)}$

{\displaystyle {\begin{aligned}\sin(a+b)=\cos({\frac {\pi }{2}}-(a+b))=\cos(({\frac {\pi }{2}}-a)-b)=\cos({\frac {\pi }{2}}-a)\cos b+\sin({\frac {\pi }{2}}-a)\sin b=\sin a\cos b+\cos a\sin b\end{aligned}}}

${\displaystyle \tan(\alpha \pm \beta )={\frac {\sin(\alpha \pm \beta )}{\cos(\alpha \pm \beta )}}={\frac {\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }{\cos \alpha \cos \beta \pm \sin \alpha \sin \beta }}={\frac {{\frac {\sin \alpha \cos \beta }{\cos \alpha \cos \beta }}\pm {\frac {\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}}{{\frac {\cos \alpha \cos \beta }{\cos \alpha \cos \beta }}\pm {\frac {\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}}}={\frac {{\frac {\sin \alpha }{\cos \alpha }}\pm {\frac {\sin \beta }{\cos \beta }}}{1\pm {\frac {\sin \alpha }{\cos \alpha }}\cdot {\frac {\sin \beta }{\cos \beta }}}}={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$

• 两角和与差的正弦公式：${\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$
• 两角和与差的余弦公式：${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$
• 两角和与差的正切公式：${\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$

2个推广到任意角的公式：

• ${\displaystyle \cos({\frac {\pi }{2}}-a)=\sin a}$
• ${\displaystyle \sin({\frac {\pi }{2}}-a)=\cos a}$

#### 正弦与余弦的和/差角公式的练习题

(1) ${\displaystyle \sin 75^{\circ }+\cos {\frac {5\pi }{12}}}$
(2) ${\displaystyle \sin 45^{\circ }\sin 15^{\circ }+\cos 45^{\circ }\cos 15^{\circ }}$
(3) ${\displaystyle \cos 44^{\circ }\sin 14^{\circ }-\sin 44^{\circ }\cos 14^{\circ }}$
(4) ${\displaystyle {\frac {\cos(x-{\frac {\pi }{4}})}{\sin x+\cos x}}}$
(5) ${\displaystyle \cos {\frac {5\pi }{12}}\cos {\frac {\pi }{6}}+\cos {\frac {\pi }{12}}\sin {\frac {\pi }{6}}}$
(6) ${\displaystyle 2(\sin 35^{\circ }\cos 25^{\circ }+\sin 55^{\circ }\cos 65^{\circ })}$
(7) ${\displaystyle {\frac {2\cos 10^{\circ }-\sin 20^{\circ }}{\cos 20^{\circ }}}}$

(1)
${\displaystyle {\begin{array}{l}\sin 75^{\circ }+\cos {\frac {5\pi }{12}}\\=\sin 75^{\circ }+\cos 75^{\circ }\\=\sin(45^{\circ }+30^{\circ })+\cos(45^{\circ }+30^{\circ })\\=(\sin 45^{\circ }\cos 30^{\circ }+\cos 45^{\circ }\sin 30^{\circ })+(\cos 45^{\circ }\cos 30^{\circ }-\sin 45^{\circ }\sin 30^{\circ })\\=({\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {3}}{2}}+{\frac {\sqrt {2}}{2}}\cdot {\frac {1}{2}})+({\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {3}}{2}}-{\frac {\sqrt {2}}{2}}\cdot {\frac {1}{2}})\\=({\frac {\sqrt {6}}{4}}+{\frac {\sqrt {2}}{4}})+({\frac {\sqrt {6}}{4}}-{\frac {\sqrt {2}}{4}})\\={\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}+{\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}\\={\frac {2{\sqrt {6}}}{4}}={\frac {\sqrt {6}}{2}}\end{array}}}$
(2)
${\displaystyle {\begin{array}{l}\sin 45^{\circ }\sin 15^{\circ }+\cos 45^{\circ }\cos 15^{\circ }\\=\cos(45^{\circ }-15^{\circ })=\cos 30^{\circ }={\frac {\sqrt {3}}{2}}\end{array}}}$
(3)
${\displaystyle {\begin{array}{l}\cos 44^{\circ }\sin 14^{\circ }-\sin 44^{\circ }\cos 14^{\circ }\\=\sin(14^{\circ }-44^{\circ })=-\sin(44^{\circ }-14^{\circ })\\=-\sin 30^{\circ }=-{\frac {1}{2}}\end{array}}}$
(4)
${\displaystyle {\begin{array}{l}{\frac {\cos(x-{\frac {\pi }{4}})}{\sin x+\cos x}}\\={\frac {\cos x\cos {\frac {\pi }{4}}+\sin x\sin {\frac {\pi }{4}}}{\sin x+\cos x}}\\={\frac {\cos x\cdot {\frac {\sqrt {2}}{2}}+\sin x\cdot {\frac {\sqrt {2}}{2}}}{\sin x+\cos x}}\\={\frac {\sqrt {2}}{2}}\cdot {\frac {\cos x+\sin x}{\sin x+\cos x}}\\={\frac {\sqrt {2}}{2}}\end{array}}}$
(5)
${\displaystyle {\begin{array}{l}\cos {\frac {5\pi }{12}}\cos {\frac {\pi }{6}}+\cos {\frac {\pi }{12}}\sin {\frac {\pi }{6}}\\=\cos({\frac {\pi }{2}}-{\frac {\pi }{12}})\cos {\frac {\pi }{6}}+\cos {\frac {\pi }{12}}\sin {\frac {\pi }{6}}\\=\sin {\frac {\pi }{12}}\cos {\frac {\pi }{6}}+\cos {\frac {\pi }{12}}\sin {\frac {\pi }{6}}\\=\sin({\frac {\pi }{12}}+{\frac {\pi }{6}})\\=\sin {\frac {\pi }{4}}={\frac {\sqrt {2}}{2}}\end{array}}}$
(6)
${\displaystyle {\begin{array}{l}2(\sin 35^{\circ }\cos 25^{\circ }+\sin 55^{\circ }\cos 65^{\circ })\\=2(\sin 35^{\circ }\cos 25^{\circ }+\sin(90^{\circ }-35^{\circ })\cos(90^{\circ }-25^{\circ }))\\=2(\sin 35^{\circ }\cos 25^{\circ }+\cos 35^{\circ }\sin 25^{\circ })\\=2\sin(35^{\circ }+25^{\circ })=2\sin 60^{\circ }\\=2\times {\frac {\sqrt {3}}{2}}={\sqrt {3}}\end{array}}}$
(7)
${\displaystyle {\begin{array}{l}{\frac {2\cos 10^{\circ }-\sin 20^{\circ }}{\cos 20^{\circ }}}\\={\frac {2\cos(30^{\circ }-20^{\circ })-\sin 20^{\circ }}{\cos 20^{\circ }}}\\={\frac {2(\cos 30^{\circ }\cos 20^{\circ }+\sin 30^{\circ }\sin 20^{\circ })-\sin 20^{\circ }}{\cos 20^{\circ }}}\\={\frac {2({\frac {\sqrt {3}}{2}}\cos 20^{\circ }+{\frac {1}{2}}\sin 20^{\circ })-\sin 20^{\circ }}{\cos 20^{\circ }}}\\={\frac {{\sqrt {3}}\cos 20^{\circ }+\sin 20^{\circ }-\sin 20^{\circ }}{\cos 20^{\circ }}}\\={\frac {{\sqrt {3}}\cos 20^{\circ }}{\cos 20^{\circ }}}={\sqrt {3}}\end{array}}}$

(1) 求${\displaystyle \cos(\alpha -{\frac {\pi }{3}})}$的值。
(2) 求${\displaystyle \sin \beta }$的值。

#### 涉及正切的和/差角公式的练习题

(1) ${\displaystyle {\frac {\tan 80^{\circ }-\tan 20^{\circ }}{1+\tan 80^{\circ }\tan 20^{\circ }}}}$
(2) ${\displaystyle {\frac {{\sqrt {3}}+\tan 15^{\circ }}{1-{\sqrt {3}}\tan 15^{\circ }}}}$
(3) ${\displaystyle {\frac {1+\tan 15^{\circ }}{1-\tan 15^{\circ }}}}$
(4) ${\displaystyle \tan 25^{\circ }+\tan 35^{\circ }+{\sqrt {3}}\tan 25^{\circ }\tan 35^{\circ }}$
(5) ${\displaystyle \tan 36^{\circ }+\tan 84^{\circ }-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }}$
(6) ${\displaystyle (\tan 10^{\circ }-{\sqrt {3}}){\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}}$

(1)
${\displaystyle {\begin{array}{l}{\frac {\tan 80^{\circ }-\tan 20^{\circ }}{1+\tan 80^{\circ }\tan 20^{\circ }}}\\=\tan(80^{\circ }-20^{\circ })=\tan 60^{\circ }={\sqrt {3}}\end{array}}}$
(2)
${\displaystyle {\begin{array}{l}{\frac {{\sqrt {3}}+\tan 15^{\circ }}{1-{\sqrt {3}}\tan 15^{\circ }}}={\frac {\tan 60^{\circ }+\tan 15^{\circ }}{1-\tan 60^{\circ }\tan 15^{\circ }}}\\=\tan(60^{\circ }+15^{\circ })=\tan(45^{\circ }+30^{\circ })\\={\frac {\tan 45^{\circ }+\tan 30^{\circ }}{1-\tan 45^{\circ }\tan 30^{\circ }}}\\={\frac {1+{\frac {\sqrt {3}}{3}}}{1-1\times {\frac {\sqrt {3}}{3}}}}={\frac {\frac {3+{\sqrt {3}}}{3}}{\frac {3-{\sqrt {3}}}{3}}}={\frac {3+{\sqrt {3}}}{3-{\sqrt {3}}}}\\={\frac {(3+{\sqrt {3}})(3+{\sqrt {3}})}{(3-{\sqrt {3}})(3+{\sqrt {3}})}}={\frac {(3+{\sqrt {3}})^{2}}{3^{2}-({\sqrt {3}})^{2}}}={\frac {9+6{\sqrt {3}}+3}{9-3}}\\={\frac {12+6{\sqrt {3}}}{6}}=2+{\sqrt {3}}\end{array}}}$
(3)
${\displaystyle {\begin{array}{l}{\frac {1+\tan 15^{\circ }}{1-\tan 15^{\circ }}}={\frac {1+\tan 15^{\circ }}{1-1\times \tan 15^{\circ }}}\\={\frac {\tan 45^{\circ }+\tan 15^{\circ }}{1-\tan 45^{\circ }\tan 15^{\circ }}}\\=\tan(45^{\circ }+15^{\circ })=\tan 60^{\circ }={\sqrt {3}}\end{array}}}$
(4)
${\displaystyle \tan(25^{\circ }+35^{\circ })={\frac {\tan 25^{\circ }+\tan 35^{\circ }}{1-\tan 25^{\circ }\tan 35^{\circ }}}\quad \Leftrightarrow \quad \tan 25^{\circ }+\tan 35^{\circ }=\tan(25^{\circ }+35^{\circ })(1-\tan 25^{\circ }\tan 35^{\circ })}$

${\displaystyle {\begin{array}{l}(\tan 25^{\circ }+\tan 35^{\circ })+{\sqrt {3}}\tan 25^{\circ }\tan 35^{\circ }\\=\tan(25^{\circ }+35^{\circ })(1-\tan 25^{\circ }\tan 35^{\circ })+{\sqrt {3}}\tan 25^{\circ }\tan 35^{\circ }\\=\tan 60^{\circ }(1-\tan 25^{\circ }\tan 35^{\circ })+{\sqrt {3}}\tan 25^{\circ }\tan 35^{\circ }\\={\sqrt {3}}(1-\tan 25^{\circ }\tan 35^{\circ })+{\sqrt {3}}\tan 25^{\circ }\tan 35^{\circ }\\={\sqrt {3}}(1-\tan 25^{\circ }\tan 35^{\circ }+\tan 25^{\circ }\tan 35^{\circ })\\={\sqrt {3}}\end{array}}}$
(5)
${\displaystyle \tan(36^{\circ }+84^{\circ })={\frac {\tan 36^{\circ }+\tan 84^{\circ }}{1-\tan 36^{\circ }\tan 84^{\circ }}}\quad \Leftrightarrow \quad \tan 36^{\circ }+\tan 84^{\circ }=\tan(36^{\circ }+84^{\circ })(1-\tan 36^{\circ }\tan 84^{\circ })}$

${\displaystyle {\begin{array}{l}(\tan 36^{\circ }+\tan 84^{\circ })-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=\tan(36^{\circ }+84^{\circ })(1-\tan 36^{\circ }\tan 84^{\circ })-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=\tan 120^{\circ }(1-\tan 36^{\circ }\tan 84^{\circ })-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=\tan(-60^{\circ })(1-\tan 36^{\circ }\tan 84^{\circ })-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=(-{\sqrt {3}})(1-\tan 36^{\circ }\tan 84^{\circ })-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=-{\sqrt {3}}+{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }-{\sqrt {3}}\tan 36^{\circ }\tan 84^{\circ }\\=-{\sqrt {3}}\end{array}}}$
(6)
${\displaystyle \tan(10^{\circ }-60^{\circ })={\frac {\tan 10^{\circ }-\tan 60^{\circ }}{1+\tan 10^{\circ }\tan 60^{\circ }}}\quad \Leftrightarrow \quad \tan 10^{\circ }-\tan 60^{\circ }=\tan(10^{\circ }-60^{\circ })(1+\tan 10^{\circ }\tan 60^{\circ })}$

${\displaystyle {\begin{array}{l}(\tan 10^{\circ }-{\sqrt {3}}){\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\=(\tan 10^{\circ }-\tan 60^{\circ }){\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\=({\frac {\sin 10^{\circ }}{\cos 10^{\circ }}}-{\frac {\sin 60^{\circ }}{\cos 60^{\circ }}}){\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\=({\frac {\sin 10^{\circ }\cos 60^{\circ }-\cos 10^{\circ }\sin 60^{\circ }}{\cos 10^{\circ }\cos 60^{\circ }}}){\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\={\frac {\sin(10^{\circ }-60^{\circ })}{\cos 10^{\circ }\cos 60^{\circ }}}\cdot {\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\={\frac {\sin(-50^{\circ })}{\cos 10^{\circ }\cos 60^{\circ }}}\cdot {\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\={\frac {-\sin 50^{\circ }}{\cos 10^{\circ }\cos 60^{\circ }}}\cdot {\frac {\cos 10^{\circ }}{\sin 50^{\circ }}}\\={\frac {-1}{\cos 60^{\circ }}}={\frac {-1}{\frac {1}{2}}}=-2\end{array}}}$

### 诱导公式

• ${\displaystyle \sin(\pi -\alpha )=\sin \alpha }$
• ${\displaystyle \cos(\pi -\alpha )=-\cos \alpha }$
• ${\displaystyle \sin \left({\frac {\pi }{2}}-\alpha \right)=\cos \alpha }$
• ${\displaystyle \cos \left({\frac {\pi }{2}}-\alpha \right)=\sin \alpha }$
• ${\displaystyle \tan \left({\frac {\pi }{2}}-\alpha \right)=\cot \alpha }$

(1) ${\displaystyle \cos 15^{\circ }\cos 105^{\circ }+\sin 15^{\circ }\sin 105^{\circ }}$
(2) ${\displaystyle \sin 460^{\circ }\sin(-160^{\circ })+\cos 560^{\circ }\cos(-280^{\circ })}$
(3) ${\displaystyle \sin 125^{\circ }\sin 245^{\circ }+\sin 35^{\circ }\sin 155^{\circ }}$

(1) ${\displaystyle \cos {\frac {5\pi }{12}}\cos {\frac {\pi }{6}}+\cos {\frac {\pi }{12}}\sin {\frac {\pi }{6}}}$
(2) ${\displaystyle \sin 14^{\circ }\cos 16^{\circ }+\sin 76^{\circ }\cos 74^{\circ }}$

## 常用结论与常见模型

### 角的配凑

• ${\displaystyle a=(a-b)+b}$
• ${\displaystyle 2a=(a+b)+(a-b)}$
• ${\displaystyle a={\frac {a+b}{2}}+{\frac {a-b}{2}}}$
• ${\displaystyle 2b=(a+b)-(a-b)}$

(1) ${\displaystyle \cos(x-45^{\circ })\cos(x+15^{\circ })+\sin(x-45^{\circ })\sin(x+15^{\circ })}$
(2) ${\displaystyle \sin(x+27^{\circ })\cos(18^{\circ }-x)-\sin(63^{\circ }-x)\sin(x-18^{\circ })}$
(3) ${\displaystyle \sin(54^{\circ }-x)\cos(36^{\circ }+x)+\cos(54^{\circ }-x)\sin(36^{\circ }+x)}$
(4) ${\displaystyle {\frac {\cos 7^{\circ }-\sin 15^{\circ }\sin 8^{\circ }}{\cos 8^{\circ }}}}$

${\displaystyle \alpha +\beta ={\frac {5\pi }{4}}}$，可知${\displaystyle \tan(\alpha +\beta )=\tan({\frac {5\pi }{4}})=\tan(\pi +{\frac {\pi }{4}})=\tan({\frac {\pi }{4}})=1}$

${\displaystyle {\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}=1\quad \Leftrightarrow \quad \tan \alpha +\tan \beta =1-\tan \alpha \tan \beta \quad \Leftrightarrow \quad \tan \alpha \tan \beta =1-\tan \alpha -\tan \beta }$

${\displaystyle (1+\tan \alpha )(1+\tan \beta )=1+\tan \alpha +\tan \beta +\tan \alpha \tan \beta =1+\tan \alpha +\tan \beta +(1-\tan \alpha -\tan \beta )=2}$

(1) 求${\displaystyle \tan(\alpha +\beta -{\frac {\pi }{4}})}$的值。
(2) 求${\displaystyle \tan(\alpha +\beta )}$的值。

(1) 求${\displaystyle \tan \beta }$的值。
(2) 求${\displaystyle \alpha +\beta }$的值。

### 涉及三角形的问题

A.锐角三角形；B.直角三角形；C.等腰三角形；D.等腰三角形或直角三角形

A.锐角三角形；B.直角三角形；C.钝角三角形；D.等腰三角形

## 补充习题

### 基础与中档练习

• 下列式子中，恒等成立的是(    )。
A.${\displaystyle \sin(\alpha +\beta )=\sin \alpha +\sin \beta }$
B.${\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }$
C.${\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha -\tan \beta }{1-\tan \alpha \tan \beta }}}$
D.${\displaystyle \sin(\alpha +\beta )\sin(\alpha -\beta )=\sin ^{2}\alpha -\sin ^{2}\beta }$
• 计算${\displaystyle \cos(x-35^{\circ })\cos(x+25^{\circ })+\sin(x-35^{\circ })\sin(x+25^{\circ })}$
• 已知${\displaystyle 0<\beta <\alpha <{\frac {\pi }{2}},\sin \alpha \sin({\frac {\pi }{2}}-\beta )+\cos \alpha \cos({\frac {\pi }{2}}+\beta )={\frac {3{\sqrt {3}}}{14}}}$，点${\displaystyle P(1,4{\sqrt {3}})}$为角${\displaystyle \alpha }$终边上的一点，求${\displaystyle \beta }$的大小。
• 已知${\displaystyle \alpha \in ({\frac {\pi }{2}},\pi ),\beta \in (-{\frac {\pi }{2}},0),\sin \alpha ={\frac {12}{13}},\cos \beta ={\frac {3}{5}}}$，求${\displaystyle \cos(\alpha -\beta )}$的值。
• 已知${\displaystyle \alpha \in (0,{\frac {\pi }{4}}),\cos(\alpha -{\frac {\pi }{4}})={\frac {3}{5}}}$，分别求${\displaystyle \sin(\alpha -{\frac {\alpha }{4}})}$${\displaystyle \sin \alpha }$的值。
• 已知${\displaystyle -{\frac {\pi }{6}}<\alpha <{\frac {\pi }{3}},\sin(\alpha +{\frac {\pi }{6}})={\frac {4}{5}}}$，分别求${\displaystyle \cos(\alpha -{\frac {\pi }{3}})}$${\displaystyle \cos \alpha }$的值。
• 在平面直角坐标系xOy中，角A和B的的顶点与原点重合，始边与x轴的非负半轴重合，终边分别与单位圆交于${\displaystyle A({\frac {\sqrt {5}}{5}},{\frac {2{\sqrt {5}}}{5}}),B({\frac {-7{\sqrt {2}}}{10}},{\frac {\sqrt {2}}{10}})}$两点。
(1) 求${\displaystyle \cos(\alpha +\beta )}$的值。
(2) 限定${\displaystyle \alpha \in (0,{\frac {\pi }{2}}),\beta \in ({\frac {\pi }{2}},\pi )}$，求${\displaystyle 2\alpha -\beta }$的值。

### 提高与拓展问题

• 已知${\displaystyle \sin \alpha -\sin \beta =1-{\frac {\sqrt {3}}{2}},\cos \alpha -\cos \beta ={\frac {1}{2}}}$，求${\displaystyle \cos(\alpha -\beta )}$的值。
• 已知在三角形ABC中，${\displaystyle \cos A\cos B=k+\sin A\sin B\quad (k\in \mathbb {R} )}$。如果将k取为a，则此时C是锐角；如果将k取为b，则此时C是直角；如果将k取为c，则此时C是钝角。求a、b、c的大小关系。
• 利用两角和与差的正弦、余弦、正切公式，分别推导其它3个三角函数（余切正割余割）的两角和与两角差公式。
• 求证：${\displaystyle \tan(\theta _{1}+\theta _{2}+\theta _{3})={\frac {(\tan \theta _{1}+\tan \theta _{2}+\tan \theta _{3})-\tan \theta _{1}\tan \theta _{2}\tan \theta _{3}}{1-(\tan \theta _{1}\tan \theta _{2}+\tan \theta _{2}\tan \theta _{3}+\tan \theta _{3}\tan \theta _{1})}}}$
• 已知${\displaystyle x,y,z,A,B,C\in \mathbb {R} }$，通过展开并化简恒等式${\displaystyle (x-y\cos C-z\cos B)^{2}+(y\sin C-z\sin B)^{2}\geqslant 0}$，证明三角形内角的嵌入不等式${\displaystyle x^{2}+y^{2}+z^{2}\geqslant 2xy\cos C+2xz\cos B+2yz\cos A}$

## 参考资料

1. （英文）Tangent Identities．CliffsNotes．
2. 人民教育出版社中学数学室. 第4章“三角函数”第2部分“两角和与差的三角函数”第4.6节“两角和与差的正弦、余弦、正切”. 数学. 全日制普通高级中学教科书 (必修). 第1册 (下) 1. 中国北京沙滩后街55号: 人民教育出版社. 2003: 34–42. ISBN 7-107-17105-4 （中文（中国大陆））.