`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...
`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.
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