Expert: Ahmed Salami - 10/31/2008

Question
I WANT YOU PLEASE TO PROVE THAT  COSHAX+SINHAX=e^AX
AND COSHAX-SINHAX=e^AX

Answer
Hi Eslam,
By definition,
sinh ax = (e^ax - e^-ax)/2
and
cosh ax = (e^ax + e^-ax)/2
So,
cosh ax + sinh ax = (e^ax + e^-ax)/2 + (e^ax - e^-ax)/2
                 = e^ax/2 + e^-ax/2 + e^ax/2 - e^-ax/2
                 = e^ax/2 + e^ax/2
                 = e^ax
In the same way,
cosh ax - sinh ax = (e^ax + e^-ax)/2 - (e^ax - e^-ax)/2
                 = e^ax/2 + e^-ax/2 - e^ax/2 + e^-ax/2
                 = e^-ax/2 + e^-ax/2
                 = e^-ax

Regards