How do you prove $sinh x + cosh x = e^{x}$ $sinhx +coshx=e^x$ ?

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Answer 1 · 2016-05-04T14:54:32

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Use the definition $cosh x = \frac{e^{x} + e^{- x}}{2} and sinh x = \frac{e^{x} - e^{- x}}{2}$ $cosh x = (e^x+e^-x)/2 and sinh x = (e^x-e^-x)/2$

Left Side: $= cosh x + sinh x$ $=cosh x + sinh x$

$= \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2}$ $=(e^x+e^-x)/2 +(e^x-e^-x)/2$

$= \frac{e^{x} + e^{- x} + e^{x} - e^{- x}}{2}$ $=(e^x+e^-x +e^x-e^-x)/2$

$= \frac{e^{x} + e^{x}}{2}$ $=(e^x +e^x)/2$

$= \frac{2 e^{x}}{2}$ $=(2e^x)/2$

$= e^{x}$ $=e^x$

How do you prove sinhx +coshx=e^xsinhx+coshx=exsinhx +coshx=e^x?