$r=\sqrt{x^{2}+y^{2}}$ $\begin{array}{|c|c|} \hline \sin\theta&\cos\theta&\tan\theta&\cot\theta\\ \hline y/r& x/r&y/x&x/y \\ \hline \end{array}$

同角三角函数

$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}, \cot \alpha=\frac{\cos \alpha}{\sin \alpha}$ $\boxed{\sin ^{2} \alpha+\cos ^{2} \alpha=1}$

两角

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

倍角

$\sin 2 \alpha=2 \sin \alpha \cos \alpha \\ \begin{aligned} \cos 2 \alpha&=\cos ^{2} \alpha-\sin ^{2} \alpha\\&=2 \cos ^{2} \alpha-1\\&=1-2 \sin ^{2} \alpha\end{aligned}$

半角

$\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}$ $\sin ^{2} \alpha=\frac{1+\sin 2 \alpha}{2}=\frac{1-\cos 2 \alpha}{2}$ $\begin{array}{l} \tan \alpha&=\frac{\sin \alpha}{\cos \alpha}\\& =\frac{\sin^2\alpha}{\cos \alpha\sin \alpha}=\frac{1-\cos 2 \alpha}{\sin 2 \alpha}\\& =\frac{\sin\alpha\cos\alpha}{\cos^2 \alpha} =\frac{\sin 2 \alpha}{1+\cos 2\alpha}\end{array}$ ${\begin{aligned}\tan \frac {\theta }{2}&=\pm \,{\sqrt {1-\cos \theta \over 1+\cos \theta }}\\&=\frac {\sin \theta }{1+\cos \theta }\\&=\frac {1-\cos \theta }{\sin \theta }\\&=\frac {\cos \theta +\sin \theta -1}{\cos \theta -\sin \theta +1}\end{aligned}}$

万能

$\boxed{\sin 2 \alpha=\frac{2 \tan \alpha}{1+\tan ^{2} \alpha}, \quad \cos 2 \alpha=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}, \quad \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha} }$ $\sin \alpha=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}=\frac{2 \tan \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}=\frac{2 t}{1+t^{2}}\\ \cos \alpha=\frac{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}}=\frac{1-\tan ^{2} \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}=\frac{1-t^{2}}{1+t^{2}}\\\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{2 \tan \frac{\alpha}{2}}{1-\tan ^{2} \frac{\alpha}{2}}=\frac{2 t}{1-t^{2}}$

和差化积

$\begin{array}{l} \sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\\ \sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\\ \cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\\ \cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2} \end{array}$

积化和差

$\begin{array}{l} \sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ \cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \\ \cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha-\beta)+\cos (\alpha+\beta)] \\ \sin \alpha \sin \beta=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \end{array}$

辅助角

$a \sin \alpha \pm b \cos \alpha=\sqrt{a^{2}+b^{2}} \sin (\alpha \pm \varphi) \text { 其中 } \tan \varphi=\frac{b}{a}(a \neq 0)\\ \sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}, \cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}$

解三角形

$\cos C = \frac{a^2} + {b^2} - {c^2}{2ab} = \frac{\overrightarrow a \cdot \overrightarrow b }{|a| \cdot |b|}\\ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\\ {S_{\triangle} = \frac{1}{2}xh = \frac{1}{2}ab\sin C =\frac{1}{2}Cr}$ $\tan \alpha=\frac{\sin \alpha}{\cos \alpha}, \cot \alpha=\frac{\cos \alpha}{\sin \alpha}\\ \\\sin 2 \alpha=2 \sin \alpha \cos \alpha \\\boxed{\sin ^{2} \alpha+\cos ^{2} \alpha=1}\\ \begin{aligned} \cos 2 \alpha&=\cos ^{2} \alpha-\sin ^{2} \alpha\\&=2 \cos ^{2} \alpha-1\\&=1-2 \sin ^{2} \alpha\end{aligned}$