加法定理【三角関数】

座標平面上の原点周りの角 \(\theta\) の回転は \( \left( \begin{array}{c} x^{\prime} \\ y^{\prime} \end{array}\right) = \left( \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left( \begin{array}{c} x \\ y \end{array}\right)\) で表される.

\(R(\theta)=\left( \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\)とおくと \(R(A+B) = R(A)R(B)\) なので

\[\begin{eqnarray*} \left( \begin{array}{cc} \cos(A+B) & -\sin(A+B) \\ \sin(A+B) & \cos(A+B) \end{array}\right) = \left( \begin{array}{cc} \cos A & -\sin A \\ \sin A & \cos A \end{array}\right)\left( \begin{array}{cc} \cos B & -\sin B \\ \sin B & \cos B \end{array}\right) \\ = \left( \begin{array}{cc} \cos A\cos B -\sin A\sin B & -\sin A\cos B – \sin B\cos A \\ \sin A\cos B + \sin B\cos A & \cos A\cos B -\sin A\sin B \end{array}\right) \end{eqnarray*}\]

ゆえに

\[ \sin(A+B) = \sin A\cos B+\sin B\cos A \\ \cos(A+B) = \cos A\cos B-\sin A\sin B\]

ここで \(\sin (-x) = -\sin x, \cos(-x)=\cos x\) より

\[ \sin(A-B) = \sin A\cos B-\sin B\cos A \\ \cos(A-B) = \cos A\cos B+\sin A\sin B\]

また

\[\begin{eqnarray*} \tan(A+B) &=& \frac{\sin(A+B)}{\cos(A+B)} \\ &=&\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B-\sin A\sin B} \\ &=&\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{1-\frac{\sin A\sin B}{\cos A\cos B}} \\ &=& \frac{\tan A+\tan B}{1-\tan A\tan B} \end{eqnarray*}\]

であり \(\tan (-x) = -\tan x \)なので

\[\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\]