# 高中数学/函数与三角/万能公式与多倍角相关公式

## 基础知识

### 万能公式

${\displaystyle \sin 2\alpha ={\frac {2\tan \alpha }{1+\tan ^{2}\alpha }}}$

${\displaystyle \cos 2\alpha ={\frac {1-\tan ^{2}\alpha }{1+\tan ^{2}\alpha }}}$

${\displaystyle \tan 2\alpha ={\frac {2\tan \alpha }{1-\tan ^{2}\alpha }}}$

### 三倍角的正弦、余弦、正切公式

{\displaystyle {\begin{aligned}\sin(3x)&=\sin(2x+x)\\&=\sin(2x)\cos x+\cos(2x)\sin x\\&=(2\sin x\cos x)\cos x+(1-2\sin ^{2}x)\sin x\\&=2\sin x\cos ^{2}x+\sin x-2\sin ^{3}x\\&=2\sin x(1-sin^{2}x)+\sin x-2\sin ^{3}x\\&=2\sin x-2\sin ^{3}x+\sin x-2\sin ^{3}x\\&=3\sin x-4\sin ^{3}x\end{aligned}}}

{\displaystyle {\begin{aligned}\cos(3x)&=\cos(2x+x)\\&=\cos(2x)\cos x-\sin(2x)\sin x\\&=(1-2\sin ^{2}x)\cos x-(2\sin x\cos x)\sin x\\&=\cos x-2\sin ^{2}x\cos x-2\sin ^{2}x\cos x\\&=\cos x-4\sin ^{2}x\cos x\\&=\cos x-4(1-\cos ^{2}x)\cos x\\&=\cos x-4\cos x+4\cos ^{2}x\cos x\\&=4\cos ^{3}x-3\cos x\end{aligned}}}

{\displaystyle {\begin{aligned}\tan(3x)&=\tan(2x+x)\\&={\frac {\tan(2x)+\tan x}{1-\tan(2x)\tan x}}\\&={\frac {{\frac {2\tan x}{1-\tan ^{2}x}}+\tan x}{1-{\frac {2\tan x}{1-tan^{2}x}}\tan x}}\\&={\frac {\frac {2\tan x+(1-\tan ^{2}x)\tan x}{1-tan^{x}}}{\frac {(1-tan^{2}x)-2\tan x\tan x}{1-tan^{x}}}}\\&={\frac {2\tan x+\tan x-\tan ^{2}x\tan x}{1-tan^{2}x-2\tan ^{2}x}}\\&={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}\end{aligned}}}

{\displaystyle {\begin{aligned}\sin(3x)&=3\sin x-4\sin ^{3}x\\&=4(\sin x)({\frac {3}{4}}-\sin ^{2}x)\\&=4\sin x(({\frac {\sqrt {3}}{2}})^{2}-\sin ^{2}x)\\&=4\sin x(\sin ^{2}{\frac {\pi }{3}}-\sin ^{2}x)\\&=4\sin x(\sin {\frac {\pi }{3}}+\sin x)(\sin {\frac {\pi }{3}}-\sin x)\\&=4\sin x\cdot 2\sin({\frac {\pi }{6}}+{\frac {x}{2}})\cos({\frac {\pi }{6}}-{\frac {x}{2}})\cdot 2\sin({\frac {\pi }{6}}-{\frac {x}{2}})\cos({\frac {\pi }{6}}+{\frac {x}{2}})\\&=4\sin x\cdot 2\sin({\frac {\pi }{6}}+{\frac {x}{2}})\cos({\frac {\pi }{6}}+{\frac {x}{2}})\cdot 2\sin({\frac {\pi }{6}}-{\frac {x}{2}})\cos({\frac {\pi }{6}}-{\frac {x}{2}})\\&=4\sin x\cdot \sin(2({\frac {\pi }{6}}+{\frac {x}{2}}))\cdot \sin(2({\frac {\pi }{6}}-{\frac {x}{2}}))\\&=4\sin x\sin({\frac {\pi }{3}}+x)\sin({\frac {\pi }{3}}-x)\\\end{aligned}}}

{\displaystyle {\begin{aligned}\cos(3x)&=4\cos ^{3}x-3\cos x\\&=4\cos x(\cos ^{2}x-{\frac {3}{4}})\\&=4\cos x(\cos ^{2}x-({\frac {\sqrt {3}}{2}})^{2})\\&=4\cos x(\cos ^{2}x-\cos ^{2}{\frac {\pi }{6}})\\&=4\cos x(\cos x+\cos {\frac {\pi }{6}})(\cos x-\cos {\frac {\pi }{6}})\\&=4\cos x\cdot 2\cos({\frac {x}{2}}+{\frac {\pi }{12}})\cos({\frac {x}{2}}-{\frac {\pi }{12}})\cdot (-2\sin({\frac {x}{2}}+{\frac {\pi }{12}})\sin({\frac {x}{2}}-{\frac {\pi }{12}}))\\&=4\cos x\cdot 2\sin({\frac {x}{2}}+{\frac {\pi }{12}})\cos({\frac {x}{2}}+{\frac {\pi }{12}})\cdot (-2\sin({\frac {x}{2}}-{\frac {\pi }{12}})\cos({\frac {x}{2}}-{\frac {\pi }{12}}))\\&=4\cos x\cdot \sin(2({\frac {x}{2}}+{\frac {\pi }{12}}))\cdot (-\sin(2({\frac {x}{2}}-{\frac {\pi }{12}})))\\&=4\cos x\cdot \sin(x+{\frac {\pi }{6}})\cdot (-\sin(x-{\frac {\pi }{6}}))\\&=4\cos x\sin({\frac {\pi }{6}}+x)\sin({\frac {\pi }{6}}-x)\\&=4\cos x\cos({\frac {\pi }{2}}-({\frac {\pi }{6}}+x))\cos({\frac {\pi }{2}}-({\frac {\pi }{6}}-x))\\&=4\cos x\cos({\frac {\pi }{3}}-x)\cos({\frac {\pi }{3}}+x)\end{aligned}}}

{\displaystyle {\begin{aligned}\tan 3x&={\frac {\sin 3x}{\cos 3x}}\\&={\frac {4\sin x\cdot \sin({\frac {\pi }{3}}+x)\cdot \sin({\frac {\pi }{3}}-x)}{4\cos x\cos({\frac {\pi }{3}}-x)\cos({\frac {\pi }{3}}+x)}}\\&={\frac {\sin x}{\cos x}}\cdot {\frac {\sin({\frac {\pi }{3}}+x)}{\cos({\frac {\pi }{3}}+x)}}\cdot {\frac {\sin({\frac {\pi }{3}}-x)}{\cos({\frac {\pi }{3}}-x)}}\\&=\tan x\tan({\frac {\pi }{3}}+x)\tan({\frac {\pi }{3}}-x)\end{aligned}}}

三倍角的常见公式列举如下：

• ${\displaystyle \sin(3x)=3\sin x-4\sin ^{3}x=4\sin x\sin({\frac {\pi }{3}}+x)\sin({\frac {\pi }{3}}-x)}$
• ${\displaystyle \cos(3x)=4\cos ^{3}x-3\cos x=4\cos x\cos({\frac {\pi }{3}}-x)\cos({\frac {\pi }{3}}+x)}$
• ${\displaystyle \tan 3x=\tan x\tan({\frac {\pi }{3}}+x)\tan({\frac {\pi }{3}}-x)}$

相关例题：在三角形ABC中，角A、B、C的对边长度分别是a、b、c。已知${\displaystyle a=32,b={\sqrt {2}},A=2B}$ ，求c的值。

${\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}\quad \Rightarrow \quad {\frac {32}{\sin 2B}}={\frac {16{\sqrt {2}}}{\sin B}}\quad \Rightarrow \quad {\frac {32}{2\sin B\cos B}}={\frac {16{\sqrt {2}}}{\sin B}}\quad \Rightarrow \quad \cos B={\frac {\sqrt {2}}{2}}}$

${\displaystyle {\frac {c}{\sin C}}={\frac {b}{\sin B}}\quad \Rightarrow \quad c={\frac {b\sin C}{\sin B}}={\frac {32\times \sin(A+B)}{2}}=16\sin(3B)}$

${\displaystyle c=16\sin(3B)=16(3\sin B-4\sin ^{3}B)=16(3\times {\frac {\sqrt {2}}{2}}-4({\frac {\sqrt {2}}{2}})^{3})=8{\sqrt {2}}}$

### 多倍角公式简介

n倍角的正弦、余弦公式是比较繁琐的求和式，需要使用二项式系数表示，求和时还需要区分奇数情形与偶数情形：

{\displaystyle {\begin{aligned}\sin(n\theta )&=\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta \\\cos(n\theta )&=\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}\theta \sin ^{k}\theta \end{aligned}}}

${\displaystyle \cos(nx)=2\cdot \cos x\cdot \cos((n-1)x)-\cos((n-2)x)}$

${\displaystyle \sin(nx)=2\cdot \cos x\cdot \sin((n-1)x)-\sin((n-2)x)}$
${\displaystyle \tan(nx)={\frac {\tan((n-1)x)+\tan x}{1-\tan((n-1)x)\tan x}}}$

知识背景：${\displaystyle \cos(n\theta )}$ ${\displaystyle \cos \theta }$ 满足的函数关系${\displaystyle T_{n}(x)=f_{n}(\cos \theta )}$ 是一种以n为参数的知名多项式，叫做第一类切比雪夫多项式