三角恒等式¶

\[ \begin{align} \sin \alpha \cdot \csc \alpha &= 1 \\\\ \cos \alpha \cdot \sec \alpha &= 1 \\\\ \tan \alpha \cdot \cot \alpha &= 1 \end{align} \]

\[ \frac{\sin \alpha}{\cos \alpha} = \tan \alpha = \frac{\sec \alpha}{\csc \alpha} = \frac{1}{\cot \alpha} \]

\[ \begin{align} \sin^2 \alpha + \cos^2 \alpha &= 1 \\\\ \sec^2 \alpha - \tan^2 \alpha &= 1 \\\\ \csc^2 \alpha - \cot^2 \alpha &= 1 \end{align} \]

\[ \begin{align} \sin(2k\pi + \alpha) &= -\sin \alpha \\\\ \cos(2k\pi + \alpha) &= -\cos \alpha \\\\ \tan(2k\pi + \alpha) &= \tan \alpha \\\\ k \in Z \end{align} \]

\[ \begin{align} \sin(-\alpha) &= -\sin \alpha \\\\ \cos(-\alpha) &= \cos \alpha \\\\ \tan(-\alpha) &= -\tan \alpha \end{align} \]

\[ \begin{align} \sin(\pi + \alpha) &= -\sin \alpha \\\\ \cos(\pi + \alpha) &= -\cos \alpha \\\\ \tan(\pi + \alpha) &= \tan \alpha \end{align} \]

\[ \begin{align} \sin(\pi - \alpha) &= \sin \alpha \\\\ \cos(\pi - \alpha) &= -\cos \alpha \\\\ \tan(\pi - \alpha) &= -\tan \alpha \end{align} \]

\[ \begin{align} \sin(\frac{\pi}{2} - \alpha) &= \cos \alpha \\\\ \cos(\frac{\pi}{2} - \alpha) &= \sin \alpha \\\\ \tan(\frac{\pi}{2} - \alpha) &= \cot \alpha \end{align} \]

\[ \begin{align} \sin(\frac{\pi}{2} + \alpha) &= \cos \alpha \\\\ \cos(\frac{\pi}{2} + \alpha) &= -\sin \alpha \\\\ \tan(\frac{\pi}{2} + \alpha) &= -\cot \alpha \end{align} \]

\[ \begin{align} \sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\\\ \sin(\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \\\\ \cos(\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\\\ \cos(\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \\\\ \tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\\\ \tan(\alpha - \beta) &= \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \end{align} \]

\[ \begin{align} \sin 2\alpha &= 2\sin \alpha \cos \alpha \\\\ \cos 2\alpha &= \cos^2 \alpha - \sin^2 \alpha \\\\ &= 2\cos^2 \alpha - 1 \\\\ &= 1 - 2\sin^2 \alpha \\\\ \tan 2\alpha &= \frac{2\tan \alpha}{1 - \tan^2 \alpha} \end{align} \]

\[ \begin{align} \sin 3\alpha &= -4\sin^3 \alpha + 3\sin \alpha \\\\ \cos 3\alpha &= 4\cos^3 \alpha - 3\cos \alpha \\\\ \tan 3\alpha &= \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3\tan^2 \alpha} \end{align} \]

\[ \begin{align} \sin^2 \alpha &= \frac{1 - \cos 2\alpha}{2} \\\\ \cos^2 \alpha &= \frac{1 + \cos 2\alpha}{2} \\\\ \tan^2 \alpha &= \frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} \\\\ \tan \alpha &= \frac{\sin 2\alpha}{1 + \cos 2\alpha} \\\\ &= \frac{1 - \cos 2\alpha}{\sin 2\alpha} \end{align} \]

\[ \begin{align} 1 + \cos 2\alpha &= 2\cos^2 \alpha \\\\ 1 - \cos 2\alpha &= 2\sin^2 \alpha \end{align} \]

\[ \begin{align} \sin \alpha \cos \beta &= \frac{1}{2}(\sin(\alpha + \beta) + \sin(\alpha - \beta)) \\\\ \cos \alpha \sin \beta &= \frac{1}{2}(\sin(\alpha + \beta) - \sin(\alpha - \beta)) \\\\ \cos \alpha \cos \beta &= \frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta)) \\\\ \sin \alpha \sin \beta &= -\frac{1}{2}(\cos(\alpha + \beta) - \cos(\alpha - \beta)) \end{align} \]

\[ \begin{align} \sin \alpha + \sin \beta &= 2\sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \\\\ \sin \alpha - \sin \beta &= 2\cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \\\\ \cos \alpha + \cos \beta &= 2\cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \\\\ \cos \alpha - \cos \beta &= -2\sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \end{align} \]

\[ \begin{align} \sin \alpha &= \frac{2\tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \\\\ \cos \alpha &= \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \\\\ \tan \alpha &= \frac{2\tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} \end{align} \]

\[ \begin{align} a\sin \alpha + b\cos \alpha = \sqrt{a^2 + b^2} \sin(\alpha + \varphi) \\\\ \text{其中} \cos \varphi = \frac{a}{\sqrt{a^2 + b^2}}, \sin \varphi = \frac{b}{\sqrt{a^2 + b^2}} \end{align} \]

\[ \begin{align} \tan \alpha + \frac{1}{\tan \alpha} &= \frac{2}{\sin 2\alpha} \\\\ \tan \alpha - \frac{1}{\tan \alpha} &= -\frac{2}{\tan 2\alpha} \end{align} \]