双曲線関数の合成公式
公式
以下が成り立つ。
$$ c_{1} \cosh x + c_{2} \sinh x = \begin{cases} A\cosh(x + y_{1}) & \text{if } c_{1} \gt c_{2} \\ B e^{x} & \text{if } c_{1} = c_{2} = B \\ C\sinh(x + y_{2}) & \text{if } c_{1} \lt c_{2} \end{cases} $$
- $A = \sqrt{c_{2}^{2} - c_{1}^{2}}$
- $C = \sqrt{c_{2}^{2} - c_{1}^{2}}$
- $y_{1} = \cosh^{-1}\left( \dfrac{c_{1}}{\sqrt{c_{1}^{2} - c_{2}^{2}}} \right) = \sinh^{-1} \left( \dfrac{c_{2}}{\sqrt{c_{1}^{2} - c_{2}^{2}}} \right)$
- $y_{2} = \cosh^{-1}\left( \dfrac{c_{2}}{\sqrt{c_{2}^{2} - c_{1}^{2}}} \right) = \sinh^{-1} \left( \dfrac{c_{1}}{\sqrt{c_{2}^{2} - c_{1}^{2}}} \right)$
証明
$c_{1} \gt c_{2}$
$$ c_{1}\cosh x + c_{2}\sinh x = \sqrt{c_{1}^{2} - c_{2}^{2}}\left( \dfrac{c_{1}}{\sqrt{c_{1}^{2} - c_{2}^{2}}}\cosh x + \dfrac{c_{2}}{\sqrt{c_{1}^{2} - c_{2}^{2}}}\sinh x \right) $$
$\dfrac{c_{1}}{\sqrt{c_{1}^{2} - c_{2}^{2}}} > 1$とすると、これを$\cosh y$とすると、
$$ \sinh^{2} y = \cosh^{y} - 1 = \dfrac{c_{1}^{2}}{c_{1}^{2} - c_{2}^{2}} - \dfrac{c_{1}^{2} - c_{2}^{2}}{c_{1}^{2} - c_{2}^{2}} = \dfrac{c_{2}^{2}}{c_{1}^{2} - c_{2}^{2}} $$
$$ \implies \sinh y= \dfrac{c_{2}}{\sqrt{c_{1}^{2} - c_{2}^{2}}} $$
今、$A = \sqrt{c_{1}^{2} - c_{2}^{2}}$とすると、加法定理により、
$$ \begin{align*} c_{1}\cosh x + c_{2}\sinh x &= A(\cosh y \cosh x + \sinh y \sinh x) \\ &= A\cosh(x + y) \end{align*} $$
$c_{1} = c_{2}$
$B = c_{1} = c_{2}$とすると、双曲線関数の定義により、
$$ \begin{align*} c_{1}\cosh x + c_{2}\sinh x &= B(\cosh x + \sinh x) \\ &= B\left( \dfrac{e^{x} + e^{-x}}{2} + \dfrac{e^{x} - e^{-x}}{2} \right) \\ &= B e^{x} \end{align*} $$
$c_{1} \lt c_{2}$
$$ c_{1}\cosh x + c_{2}\sinh x = \sqrt{c_{2}^{2} - c_{1}^{2}}\left( \dfrac{c_{1}}{\sqrt{c_{2}^{2} - c_{1}^{2}}}\cosh x + \dfrac{c_{2}}{\sqrt{c_{2}^{2} - c_{1}^{2}}}\sinh x \right) $$
$\dfrac{c_{2}}{\sqrt{c_{2}^{2} - c_{1}^{2}}} > 1$とすると、これを$\cosh y$とすると、
$$ \sinh^{2} y = \cosh^{y} - 1 = \dfrac{c_{2}^{2}}{c_{2}^{2} - c_{1}^{2}} - \dfrac{c_{2}^{2} - c_{1}^{2}}{c_{2}^{2} - c_{1}^{2}} = \dfrac{c_{1}^{2}}{c_{2}^{2} - c_{1}^{2}} $$
$$ \implies \sinh y= \dfrac{c_{1}}{\sqrt{c_{2}^{2} - c_{1}^{2}}} $$
今、$C = \sqrt{c_{2}^{2} - c_{1}^{2}}$とすると、加法定理により、
$$ \begin{align*} c_{1}\cosh x + c_{2}\sinh x &= C(\sinh y \cosh x + \cosh y \sinh x) \\ &= C\sinh(x + y) \end{align*} $$
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