can anyone show me how to prove sinh(x+y)=sinhx coshy + coshx sinhy?

Answer 1

start from 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+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))]
that is left hand side= sinh(x+y).
that is it. good luck

Answer 2

What are your definitions of sinh and cosh. You may or may not know that sinhx = (e^x - e^{-x})/2 and coshx = (e^x + e^{-x})/2. It shouldn't be hard to prove the above identity with these definitions.

Answer 3

Go to the definitions:

sinh ( x ) = (e^x - e^{-x})/2
cosh( x ) = (e^x + e^{-x})/2

Plug these in for both sides, and simplify the right hand side.

Answer 4

NOOOO!!