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：

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

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

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

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

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

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

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 $\pi$ 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

$\sin(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots$

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

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

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

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

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

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

$\cos(ix)=i\cosh(x)$

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

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）