$\sinh2x\equiv2\sinh x\cdot\cosh x$

\sinh2x\equiv2\sinh x\cdot\cosh x

Proof

• We know:
  • \sinh x = \frac{e^x - e^{-x}}{2}
  • \cosh x = \frac{e^x + e^{-x}}{2}
• So:
  • 2\sinh x\cdot\cosh x
  • = 2\cdot\frac{e^x - e^{-x}}{2}\cdot\frac{e^x + e^{-x}}{2}
  • = \frac{(e^x - e^{-x})(e^x + e^{-x})}{2}
  • = \frac{e^{2x} - e^{-2x}}{2}
  • = \sinh2x
• That means we’ve shown that \sinh2x\equiv2\sinh x\cdot\cosh x!