Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 13

sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)
Take note that hyperbolic sine and hyperbolic cosine are defined by

sinh(u)=(e^u-e^(-u))/2

cosh(u)=(e^u+e^(-u))/2
Applying these formulas to the right side of the equation, it becomes
sinh(x+y) =(e^x-e^(-x))/2*(e^y+e^(-y))/2 +(e^x+e^(-x))/2*(e^y-e^(-y))/2
Multiplying the fractions, the right side turns into
sinh(x+y) = (e^(x+y)+e^(x-y) -e^(y-x)-e^(-(x+y)))/4 + (e^(x+y)-e^(x-y)+e^(y-x)-e^(-(x+y)))/4
Combining the like terms, it simplifies to
sinh(x+y) = (2e^(x+y) - 2e^(-(x+y)))/4
sinh(x+y)=(2(e^(x+y) - e^(-(x+y))))/4
sinh(x + y) = (e^(x+y) - e^(-(x+y)))/2
And, express the right side in terms of hyperbolic function again. Applying the definition of hyperbolic sine, the right side transforms to
sinh(x+y)=sinh(x+y)
This verifies that the given equation is an identity.

Therefore, sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y) is an identity.