`sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)` Verify the identity.

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`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


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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