Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 11

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.