$\cosh^2x - \sinh^2x = 1$

\cosh^2x - \sinh^2x = 1

Proof

• We know that:
  • \sinh x = \frac{e^x - e^{-x}}{2}
  • \cosh x = \frac{e^x + e^{-x}}{2}
• Square both functions:
  • \sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}
  • \cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}
• \cosh^2 x - \sinh^2 x
  • = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}
  • = \frac{e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}}4
  • = \frac44
  • =1
• So we’ve shown that \cosh^2x - \sinh^2x = 1!