Sunday 27 July 2014

`e^(2x) = sinh(2x) + cosh(2x)` Verify the identity.

`e^(2x)=sinh(2x)+cosh(2x)`


Take note that hyperbolic sine and hyperbolic cosine are defined as



  • `sinh(u) = (e^u-e^(-u))/2`

  • `cosh(u)=(e^u+e^(-u))/2`

Apply these two formulas to express the right side in exponential form.


`e^(2x)=(e^(2x)-e^(-2x))/2 + (e^(2x)+e^(-2x))/2`


Adding the two fractions, the right side simplifies to


`e^(2x) = (2e^(2x))/2`


`e^(2x)=e^(2x)`


This proves that the given equation is an identity.



Therefore,  `e^(2x)=sinh(2x)+cosh(2x)`  is an identity.

`e^(2x)=sinh(2x)+cosh(2x)`


Take note that hyperbolic sine and hyperbolic cosine are defined as



  • `sinh(u) = (e^u-e^(-u))/2`


  • `cosh(u)=(e^u+e^(-u))/2`

Apply these two formulas to express the right side in exponential form.


`e^(2x)=(e^(2x)-e^(-2x))/2 + (e^(2x)+e^(-2x))/2`


Adding the two fractions, the right side simplifies to


`e^(2x) = (2e^(2x))/2`


`e^(2x)=e^(2x)`


This proves that the given equation is an identity.



Therefore,  `e^(2x)=sinh(2x)+cosh(2x)`  is an identity.

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