目次
\begin{eqnarray} \sin(\alpha\pm\beta) &=& \sin\alpha\cos\beta\pm\cos\alpha\sin\beta \\ \cos(\alpha\pm\beta) &=& \cos\alpha\cos\beta\mp\sin\alpha\sin\beta \\ \tan(\alpha\pm\beta) &=& \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} \end{eqnarray}
\begin{eqnarray} \sin2\alpha &=& 2\sin\alpha\cos\alpha \\ \cos2\alpha &=& \cos^2\alpha-\sin^2\alpha \\ &=& 1-2\sin^2\alpha \\ &=& 2\cos^2\alpha-1 \\ \tan2\alpha &=& \frac{2\tan\alpha}{1-\tan^2\alpha} \end{eqnarray}
\begin{eqnarray} \sin^2\frac{\alpha}{2} &=& \frac{1-\cos\alpha}{2} \\ \cos^2\frac{\alpha}{2} &=& \frac{1+\cos\alpha}{2} \\ \tan^2\frac{\alpha}{2} &=& \frac{1-\cos\alpha}{1+\cos\alpha} \end{eqnarray}
\begin{eqnarray} \sin3\alpha &=& &3\sin\alpha&-4\sin^3\alpha \\ \cos3\alpha &=& -&3\cos\alpha&+4\cos^3\alpha \end{eqnarray}
\begin{eqnarray} \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{eqnarray}
\begin{eqnarray} \sin A+\sin B &=& &2&\sin\frac{A+B}{2}\cos\frac{A-B}{2} \\ \sin A-\sin B &=& &2&\cos\frac{A+B}{2}\sin\frac{A-B}{2} \\ \cos A+\cos B &=& &2&\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\ \cos A-\cos B &=& -&2&\sin\frac{A+B}{2}\sin\frac{A-B}{2} \end{eqnarray}
\begin{eqnarray} \sin(180^\circ-\theta) &=& \sin\theta \\ \cos(180^\circ-\theta) &=& -\cos\theta \\ \tan(180^\circ-\theta) &=& -\tan\theta \\ \sin(90^\circ\pm\theta) &=& \cos\theta \\ \cos(90^\circ\pm\theta) &=& \mp\sin\theta \\ \tan(90^\circ\pm\theta) &=& \mp\frac{1}{\tan\theta} \end{eqnarray}
三角形ABCにおいて、 \begin{equation} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \end{equation} ここで、a, b, cは各辺の長さ、A, B, Cは各角、Rは外接円の半径を表す。
三角形ABCにおいて、 \begin{eqnarray} a^2 &=& b^2 + c^2 - 2bc\cos A \\ b^2 &=& a^2 + c^2 - 2ac\cos B \\ c^2 &=& a^2 + b^2 - 2ab\cos C \end{eqnarray} ここで、a, b, cは各辺の長さ、A, B, Cは各角を表す。