Show-that-cosh-x-gt-sinh-x-

Question Number 178355 by Spillover last updated on 15/Oct/22

Show that

\cosh x > \sinh x

Answered by Ar Brandon last updated on 16/Oct/22

\cosh x - \sinh x = \frac{1}{2} ((e^{x} + e^{- x}) - (e^{x} - e^{- x})) = e^{- x} > 0

\Rightarrow \cosh x > \sinh x

Commented by Spillover last updated on 16/Oct/22

t h a n k s