verify the identity

**slapmaxwell1** · January 25th 2010, 05:01 AM

sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)

im not sure how to verify this, i suppose i would try to plug numbers into the equation and solve both sides. i suppose im looking for a more efficient way to verify this without having to plug and chug. thanks in advance

**Sudharaka** · January 25th 2010, 06:32 AM

Dear slapmaxwell1,

By definition, $Sinh(x+y)=\frac{e^{x+y}-e^{-x-y}}{2}$

$SinhxCoshy+CoshxSinhy=\frac{(e^x-e^{-x})(e^y+e^{-y})}{4}+\frac{(e^y-e^{-y})(e^x+e^{-x})}{4}$

By simplification, $SinhxCoshy+CoshxSinhy=\frac{e^{x+y}-e^{-x-y}}{2}$

Therefore, sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y)

Hope this helps.

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