$\cosh^2x + \sinh^2x = \cosh(2x)$

\cosh^2x + \sinh^2x = \cosh(2x)

Proof

• We know:
  • \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
  • = \frac{2e^{2x}+2e^{-2x}}4
  • = \frac{e^{2x}+e^{-2x}}2
  • = \cosh(2x)
• We’ve now shown thatt \cosh^2x + \sinh^2x = \cosh(2x)!