三角函數公式一覽

（注意：喺呢篇文章裏面，角度嘅單位係弧度。）

以下係一咋三角函數公式。

## 單位圓定義
喺笛卡兒坐標系統裏面，畫一個半徑係 r 嘅圓，對於喺圓上面嘅任何一點 (x,y)，有：

\sin\theta=\frac{y}{r}

\cos\theta=\frac{x}{r}

\tan\theta=\frac{y}{x}

\csc\theta=\frac{r}{y}

\sec\theta=\frac{r}{x}

\cot\theta=\frac{x}{y}

備注：r=\sqrt{x^2+y^2}

## 相互關係

\csc\theta=\frac{1}{\sin\theta}

\sec\theta=\frac{1}{\cos\theta}

\cot\theta=\frac{1}{\tan\theta}

\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sec\theta}{\csc\theta}

\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{\csc\theta}{\sec\theta}

\sin^2\theta+\cos^2\theta=1

1+\tan^2\theta=\sec^2\theta

1+\cot^2\theta=\csc^2\theta

## 移相公式

\sin(-x)=-\sin(x)

\cos(-x)=\cos(x)

\tan(-x)=-\tan(x)

\sec(-x)=\sec(x)

\csc(-x)=-\csc(x)

\cot(-x)=-\cot(x)

\sin(x+2n\pi)=\sin(x)\quad\forall n\in\mathbb{Z}

\cos(x+2n\pi)=\cos(x)\quad\forall n\in\mathbb{Z}

\tan(x+n\pi)=\tan(x)\quad\forall n\in\mathbb{Z}

\sec(x+2n\pi)=\sec(x)\quad\forall n\in\mathbb{Z}

\csc(x+2n\pi)=\csc(x)\quad\forall n\in\mathbb{Z}

\cot(x+n\pi)=\cot(x)\quad\forall n\in\mathbb{Z}

\sin(x+\pi)=-\sin(x)

\cos(x+\pi)=-\cos(x)

\sec(x+\pi)=-\sec(x)

\csc(x+\pi)=-\csc(x)

\sin(\pi-x)=\sin(x)

\cos(\pi-x)=-\cos(x)

\tan(\pi-x)=-\tan(x)

\sec(\pi-x)=-\sec(x)

\csc(\pi-x)=\csc(x)

\cot(\pi-x)=-\cot(x)

\sin(\frac{\pi}{2}+x)=\cos(x)

\cos(\frac{\pi}{2}+x)=-\sin(x)

\tan(\frac{\pi}{2}+x)=-\cot(x)

\sec(\frac{\pi}{2}+x)=-\csc(x)

\csc(\frac{\pi}{2}+x)=\sec(x)

\cot(\frac{\pi}{2}+x)=-\tan(x)

\sin(\frac{\pi}{2}-x)=\cos(x)

\cos(\frac{\pi}{2}-x)=\sin(x)

\tan(\frac{\pi}{2}-x)=\cot(x)

\sec(\frac{\pi}{2}-x)=\csc(x)

\csc(\frac{\pi}{2}-x)=\sec(x)

\cot(\frac{\pi}{2}-x)=\tan(x)

\sin(\frac{\pi}{4}+x)=\frac{\sqrt{2}}{2}(\sin(x)+\cos(x))

\cos(\frac{\pi}{4}+x)=\frac{\sqrt{2}}{2}(\cos(x)-\sin(x))

\tan(\frac{\pi}{4}+x)=\frac{1+\tan(x)}{1-\tan(x)}

\sec(\frac{\pi}{4}+x)=\frac{\sqrt{2}\sec(x)\csc(x)}{\csc(x)-\sec(x)}

\csc(\frac{\pi}{4}+x)=\frac{\sqrt{2}\sec(x)\csc(x)}{\csc(x)+\sec(x)}

\cot(\frac{\pi}{4}+x)=\frac{\cot(x)-1}{\cot(x)+1}

## 複角公式

\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)

\sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y)

\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)

\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)

\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}

\tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}

\sec(x+y)=\frac{\sec(x)\sec(y)\csc(x)\csc(y)}{\csc(x)\csc(y)-\sec(x)\sec(y)}

\sec(x-y)=\frac{\sec(x)\sec(y)\csc(x)\csc(y)}{\csc(x)\csc(y)+\sec(x)\sec(y)}

\csc(x+y)=\frac{\sec(x)\sec(y)\csc(x)\csc(y)}{\sec(x)\csc(y)+\csc(x)\sec(y)}

\csc(x-y)=\frac{\sec(x)\sec(y)\csc(x)\csc(y)}{\sec(x)\csc(y)-\csc(x)\sec(y)}

\cot(x+y)=\frac{\cot(x)\cot(y)-1}{\cot(x)+\cot(y)}

\cot(x-y)=\frac{\cot(x)\cot(y)+1}{\cot(y)-\cot(x)}

## 倍角公式

\sin(2x)=2\sin(x)\cos(x)

\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1=1-2\sin^2(x)

\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}

\sec(2x)=\frac{\sec^2(x)\csc^2(x)}{\csc^2(x)-\sec^2(x)}=\frac{\sec^2(x)}{2-\sec^2(x)}=\frac{\csc^2(x)}{\csc^2(x)-2}

\csc(2x)=\frac{\sec(x)\csc(x)}{2}

\cot(2x)=\frac{\cot^2(x)-1}{2\cot(x)}

\sin(3x)=3\sin(x)-4\sin^3(x)=4\sin(x)\sin(\frac{\pi}{3}-x)\sin(\frac{\pi}{3}+x)

\cos(3x)=4\cos^3(x)-3\cos(x)=4\cos(x)\cos(\frac{\pi}{3}-x)\cos(\frac{\pi}{3}+x)

\tan(3x)=\frac{\tan^3(x)-3\tan(x)}{3\tan^2(x)-1}=\tan(x)\tan(\frac{\pi}{3}-x)\tan(\frac{\pi}{3}+x)

\sec(3x)=\frac{\sec^3(x)}{4-3\sec^2(x)}

\csc(3x)=\frac{\csc^3(x)}{3\csc^2(x)-4}

\cot(3x)=\frac{\cot^3(x)-3\cot(x)}{3\cot^2(x)-1}

## 積化和差公式

\sin(x)\cos(y)=\frac{1}{2}(\sin(x+y)+\sin(x-y))

\cos(x)\sin(y)=\frac{1}{2}(\sin(x+y)-\sin(x-y))

\cos(x)\cos(y)=\frac{1}{2}(\cos(x+y)+\cos(x-y))

\sin(x)\sin(y)=\frac{1}{2}(\cos(x-y)-\cos(x+y))

## 和差化積公式

\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})

\sin(x)-\sin(y)=2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})

\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})

\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})

\tan(x)+\tan(y)=\frac{\sin(x+y)}{\cos(x)\cos(y)}

\tan(x)-\tan(y)=\frac{\sin(x-y)}{\cos(x)\cos(y)}

## 三隻角總和係 <math>\pi</math> 嘅條件恆等式
如果有三隻角 x，y 同 z，令到 x+y+z=\pi，噉就會有：

\tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z)

\cot(\frac{x}{2})+\cot(\frac{y}{2})+\cot(\frac{z}{2})=\cot(\frac{x}{2})\cot(\frac{y}{2})\cot(\frac{z}{2})

\tan(\frac{x}{2})\tan(\frac{y}{2})+\tan(\frac{x}{2})\tan(\frac{z}{2})+\tan(\frac{y}{2})\tan(\frac{z}{2})=1

\cot(x)\cot(y)+\cot(x)\cot(z)+\cot(y)\cot(z)=1

\sin(x)+\sin(y)+\sin(z)=4\cos(\frac{x}{2})\cos(\frac{y}{2})\cos(\frac{z}{2})

-\sin(x)+\sin(y)+\sin(z)=4\cos(\frac{x}{2})\sin(\frac{y}{2})\sin(\frac{z}{2})

\cos(x)+\cos(y)+\cos(z)=4\sin(\frac{x}{2})\sin(\frac{y}{2})\sin(\frac{z}{2})+1

-\cos(x)+\cos(y)+\cos(z)=4\sin(\frac{x}{2})\cos(\frac{y}{2})\cos(\frac{z}{2})-1

\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z)

-\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\cos(y)\cos(z)

\cos(2x)+\cos(2y)+\cos(2z)=-4\cos(x)\cos(y)\cos(z)-1

-\cos(2x)+\cos(2y)+\cos(2z)=-4\cos(x)\sin(y)\sin(z)+1

\sin^2(x)+\sin^2(y)+\sin^2(z)=2\cos(x)\cos(y)\cos(z)+2

-\sin^2(x)+\sin^2(y)+\sin^2(z)=2\cos(x)\sin(y)\sin(z)

\cos^2(x)+\cos^2(y)+\cos^2(z)=-2\cos(x)\cos(y)\cos(z)+1

-\cos^2(x)+\cos^2(y)+\cos^2(z)=-2\cos(x)\sin(y)\sin(z)+1

-\sin^2(2x)+\sin^2(2y)+\sin^2(2z)=-2\cos(2x)\sin(2y)\sin(2z)

-\cos^2(2x)+\cos^2(2y)+\cos^2(2z)=2\cos(2x)\sin(2y)\sin(2z)+1

\sin^2(\frac{x}{2})+\sin^2(\frac{y}{2})+\sin^2(\frac{z}{2})+2\sin(\frac{x}{2})\sin(\frac{y}{2})\sin(\frac{z}{2})=1

## 泰勒級數
\sin(x)=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots

\cos(x)=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots

\sin^{-1}(x)=\sum^{\infty}_{n=0}\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}=x+\frac{x^3}{6}+\frac{3x^5}{40}+\cdots

\cos^{-1}(x)=\frac{\pi}{2}-\sum^{\infty}_{n=0}\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}=\frac{\pi}{2}-x-\frac{x^3}{6}-\frac{3x^5}{40}-\cdots

\tan^{-1}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}x^{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots

## 複數公式
由歐拉恆等式可以得到：

(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx) （狄默夫公式）

\cos(x)=\frac{e^{ix}+e^{-ix}}{2}

\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}

\tan(x)=-\frac{i(e^{ix}-e^{-ix})}{e^{ix}+e^{-ix}}

\sec(x)=\frac{2}{e^{ix}+e^{-ix}}

\csc(x)=\frac{2i}{e^{ix}-e^{-ix}}

\cot(x)=\frac{i(e^{ix}+e^{-ix})}{e^{ix}-e^{-ix}}

\sin^{-1}(x)=-i\ln(ix+\sqrt{1-x^2})

\cos^{-1}(x)=-i\ln(x+\sqrt{x^2-1})

\tan^{-1}(x)=\frac{i}{2}\ln(\frac{i+x}{i-x})

\csc^{-1}(x)=-i\ln(\frac{i}{x}+\sqrt{1-\frac{1}{x^2}})

\sec^{-1}(x)=-i\ln(\frac{1}{x}+i\sqrt{1-\frac{1}{x^2}})

\cot^{-1}(x)=\frac{i}{2}\ln(\frac{x-i}{x+i})

### 雙曲函數

\sin(ix)=i\sinh(x)

\cos(ix)=\cosh(x)

\tan(ix)=i\tanh(x)

## 冇變量嘅恆等式
下面係一啲只有常數嘅恆等式：

\tan^{-1}(\frac{1}{3})+\tan^{-1}(\frac{1}{7})=\tan^{-1}(\frac{1}{2})

\tan^{-1}(\frac{1}{5})+\tan^{-1}(\frac{1}{8})=\tan^{-1}(\frac{1}{3})

\tan^{-1}(\frac{1}{2})+\tan^{-1}(\frac{1}{3})=\frac{\pi}{4}

2\tan^{-1}(\frac{1}{3})+\tan^{-1}(\frac{1}{7})=\frac{\pi}{4}

\tan^{-1}(\frac{1}{3})+\tan^{-1}(\frac{1}{5})+\tan^{-1}(\frac{1}{7})+\tan^{-1}(\frac{1}{8})=\frac{\pi}{4}

4\tan^{-1}(\frac{1}{5})-\tan^{-1}(\frac{1}{239})=\frac{\pi}{4}

5\tan^{-1}(\frac{1}{7})+2\tan^{-1}(\frac{3}{79})=\frac{\pi}{4}（歐拉）

\tan^{-1}{1}+\tan^{-1}{2}+\tan^{-1}{3}=\pi

\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{5}{13}+\cos^{-1}\frac{16}{65}=\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{12}{13}+\sin^{-1}\frac{63}{65}=\pi

\cos(\frac{\pi}{9})\cdot\cos(\frac{2\pi}{9})\cdot\cos(\frac{4\pi}{9})=\frac{1}{8}

\sin(\frac{\pi}{9})\cdot\sin(\frac{2\pi}{9})\cdot\sin(\frac{4\pi}{9})=\frac{\sqrt{3}}{8}

\sin(\frac{\pi}{18})\cdot\cos(\frac{5\pi}{18})\cdot\cos(\frac{7\pi}{18})=\frac{1}{8}

\sin(\frac{\pi}{12})\cdot\sin(\frac{5\pi}{12})=\frac{1}{4}

\cos(\frac{\pi}{12})\cdot\cos(\frac{5\pi}{12})=\frac{1}{4}

\tan(\frac{2\pi}{9})\cdot\tan(\frac{\pi}{6})\cdot\tan(\frac{\pi}{9})=\tan(\frac{\pi}{18})

\tan(\frac{5\pi}{18})\cdot\tan(\frac{\pi}{3})\cdot\tan(\frac{7\pi}{18})=\tan(\frac{4\pi}{9})

\cos(\frac{2\pi}{15})+\cos(\frac{4\pi}{15})+\cos(\frac{8\pi}{15})+\cos(\frac{14\pi}{15})=\frac{1}{2}

\sin^2(\frac{\pi}{10})+\sin^2(\frac{\pi}{6})=\sin^2(\frac{\pi}{5})（歐幾里得）

類:三角函數
類:一覽