`sinh(2x) = 2sinhxcoshx` Verify the identity.

`sinh(2x)=2sinh(x)cosh(x)`

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`

So when the left side is expressed in exponential form, it becomes

`(e^(2x)-e^(-2x))/2=2sinh(x)cosh(x)`

Factoring the numerator, it turns into

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

To return it to hyperbolic function, multiply the left side...

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`sinh(2x)=2sinh(x)cosh(x)`

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`

So when the left side is expressed in exponential form, it becomes

`(e^(2x)-e^(-2x))/2=2sinh(x)cosh(x)`

Factoring the numerator, it turns into

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

To return it to hyperbolic function, multiply the left side by 2/2.

`((e^x - e^(-x))(e^x + e^(-x)))/2*2/2=2sinh(x)cosh(x)`

Then, rearrange the factors in such a way that it can be expressed in terms of sinh and cosh.

`2*(e^x - e^(-x))/2 * (e^x + e^(-x))/2=2sinh(x)cosh(x)`

`2*sinh(x)*cosh(x)=2sinh(x)cosh(x)`

`2sinh(x)cosh(x)=2sinh(x)cosh(x)`

This proves that the given equation is an identity.

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

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