双曲線関数の加法定理とは
\[ \sinh(A+B) = \sinh A\cosh B+\sinh B\cosh A \\ \sinh(A-B) = \sinh A\cosh B-\sinh B\cosh A \\ \cosh(A+B) = \cosh A\cosh B+\sinh A\sinh B\\ \cosh(A-B) = \cosh A\cosh B-\sinh A\sinh B\]
\(\tanh\)の場合は
\[\tanh(A+B)=\frac{\tanh A+\tanh B}{1+\tanh A\tanh B}\\ \tanh(A-B)=\frac{\tanh A-\tanh B}{1-\tanh A\tanh B}\]
証明
\( \displaystyle \sinh x = \frac{e^x – e^{-x}}{2}, \cosh x = \frac{e^x + e^{-x}}{2} \) より
\[\begin{eqnarray*}\sinh A\cosh B+\sinh B\cosh A &=& \frac{e^A – e^{-A}}{2}\frac{e^B + e^{-B}}{2}+\frac{e^B – e^{-B}}{2}\frac{e^A + e^{-A}}{2}\\&=&\frac{e^{A+B} – e^{-(A+B)}}{2}\\&=&\sinh(A+B) \\\cosh A\cosh B+\sinh A\sinh B &=& \frac{e^A + e^{-A}}{2}\frac{e^B + e^{-B}}{2}+\frac{e^A – e^{-A}}{2}\frac{e^B – e^{-B}}{2}\\&=&\frac{e^{A+B} + e^{-(A+B)}}{2}\\&=&\cosh(A+B)\end{eqnarray*}\]
\( \sinh (-x) = -\sinh x, \cosh (-x)=\cosh x\) なので
\[ \sinh(A-B) = \sinh A\cosh B-\sinh B\cosh A \\ \cosh(A-B) = \cosh A\cosh B-\sinh A\sinh B\]
また \[ \tanh x = \frac{\sinh x}{\cosh x} \] より
\[\begin{eqnarray*}\tanh(A+B)&=&\frac{\sinh(A+B)}{\cosh(A+B)}\\ &=&\frac{\sinh A\cosh B+\sinh B\cosh A}{\cosh A\cosh B+\sinh A\sinh B}\\ &=&\frac{\frac{\sinh A}{\cosh A}+\frac{\sinh B}{\cosh B}}{1+\frac{\sinh A\sinh B}{\cosh A\cosh B}}\\ &=& \frac{\tanh A+\tanh B}{1+\tanh A\tanh B} \end{eqnarray*} \]
\( \tanh(-x) = -\tanh x\) なので
\[\tanh(A-B)=\frac{\tanh A-\tanh B}{1-\tanh A\tanh B}\]
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