# `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...

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`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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