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双曲線関数の合成公式 📂関数

双曲線関数の合成公式

公式

以下が成り立つ。

c1cosh⁡x+c2sinh⁡x={Acosh⁡(x+y1)if c1>c2Bexif c1=c2=BCsinh⁡(x+y2)if c1<c2 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=c22−c12A = \sqrt{c_{2}^{2} - c_{1}^{2}}
  • C=c22−c12C = \sqrt{c_{2}^{2} - c_{1}^{2}}
  • y1=cosh⁡−1(c1c12−c22)=sinh⁡−1(c2c12−c22)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)
  • y2=cosh⁡−1(c2c22−c12)=sinh⁡−1(c1c22−c12)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

c1>c2c_{1} \gt c_{2}

c1cosh⁡x+c2sinh⁡x=c12−c22(c1c12−c22cosh⁡x+c2c12−c22sinh⁡x) 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)

c1c12−c22>1\dfrac{c_{1}}{\sqrt{c_{1}^{2} - c_{2}^{2}}} > 1とすると、これをcosh⁡y\cosh yとすると、

sinh⁡2y=cosh⁡y−1=c12c12−c22−c12−c22c12−c22=c22c12−c22 \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}}

  ⟹  sinh⁡y=c2c12−c22 \implies \sinh y= \dfrac{c_{2}}{\sqrt{c_{1}^{2} - c_{2}^{2}}}

今、A=c12−c22A = \sqrt{c_{1}^{2} - c_{2}^{2}}とすると、加法定理により、

c1cosh⁡x+c2sinh⁡x=A(cosh⁡ycosh⁡x+sinh⁡ysinh⁡x)=Acosh⁡(x+y) \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*}

c1=c2c_{1} = c_{2}

B=c1=c2B = c_{1} = c_{2}とすると、双曲線関数の定義により、

c1cosh⁡x+c2sinh⁡x=B(cosh⁡x+sinh⁡x)=B(ex+e−x2+ex−e−x2)=Bex \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*}

c1<c2c_{1} \lt c_{2}

c1cosh⁡x+c2sinh⁡x=c22−c12(c1c22−c12cosh⁡x+c2c22−c12sinh⁡x) 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)

c2c22−c12>1\dfrac{c_{2}}{\sqrt{c_{2}^{2} - c_{1}^{2}}} > 1とすると、これをcosh⁡y\cosh yとすると、

sinh⁡2y=cosh⁡y−1=c22c22−c12−c22−c12c22−c12=c12c22−c12 \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}}

  ⟹  sinh⁡y=c1c22−c12 \implies \sinh y= \dfrac{c_{1}}{\sqrt{c_{2}^{2} - c_{1}^{2}}}

今、C=c22−c12C = \sqrt{c_{2}^{2} - c_{1}^{2}}とすると、加法定理により、

c1cosh⁡x+c2sinh⁡x=C(sinh⁡ycosh⁡x+cosh⁡ysinh⁡x)=Csinh⁡(x+y) \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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