`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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