Prove the Identity: cosh(−x)=cosh(x)\cosh(-x) = \cosh(x)

I am trying to prove this identity in my Calculus 1 class. Here is what I’ve got so far:

cosh(−x)=cosh(x)\cosh(-x) = \cosh(x)

cosh(−x)=12(e−x+e−(−x))\cosh(-x) = \frac{1}{2}(e^{-x} + e^{-{(-x)}})

cosh(−x)=12(e−x+ex)\cosh(-x)= \frac{1}{2}(e^{-x} + e^x)

Any input is much appreciated.




This looks right to me. I might change the structure of the proof, i.e., compute cosh(−x)\cosh(-x) and simplify it to cosh(x)\cosh(x), but the idea is certainly correct.
– Giuseppe
2 days ago



Just commute the two exponentials. And your done…
– Eleven-Eleven
2 days ago



Awesome. I see now. Thank you guys so much.
– J. Armstrong
2 days ago


1 Answer


You are almost there.

cosh(−x)=12(e−x+ex)=12(ex+e−x)\cosh(-x)= \frac{1}{2}(e^{-x} + e^x) = \frac{1}{2}(e^x + e^{-x} ) (commutative property)

And the RHS is the definition of cosh(x)\cosh(x)