# hyperbolic functions

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• Jan 20th 2007, 11:12 AM
dadon
hyperbolic functions
hi all,

how do you prove the following:

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

thanks.
• Jan 20th 2007, 01:32 PM
ticbol
Quote:

Originally Posted by dadon
hi all,

how do you prove the following:

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

thanks.

----sinh u = (1/2)[e^u -e^(-u)]
----cosh u = (1/2)[e^u +e^(-u)]
----a^b * a^c = a^(b+c)

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

We develop the RHS.
RHS =
= {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

= (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

Doing FOILs in the expansion,

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y) +e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

= (1/4)[2e^(x+y) -2e^(-x-y)]

= (1/2)[e^(x+y) -e^(-x-y)]

= (1/2)[e^(x+y) -e^[-(x+y)]

= sinh (x+y)

= LHS

Therefore, proven.
• Jan 20th 2007, 02:40 PM
dadon
thank you

Nicely explained!