`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...

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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

`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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