1. ## Hyperbolic functions

From the definition of sinhx and coshx, in terms of exponentials show that,

sinhx+sinhy=2sinh((x+y)/2).cosh((x-y)/2)

Can anyone help explain how to solve this? Cheers.

2. Originally Posted by Haris
From the definition of sinhx and coshx, in terms of exponentials show that,

sinhx+sinhy=2sinh((x+y)/2).cosh((x-y)/2)

Can anyone help explain how to solve this? Cheers.
$sinh(x) = \frac{e^x-e^{-x}}{2}$

$sinh(y) = \frac{e^y-e^{-y}}{2}$

$cosh(\frac{x-y}{2}) = \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}$

$sinh(\frac{x+y}{2}) = \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2}$

Hence you're trying to prove that:

$\frac{e^x-e^{-x}}{2} + \frac{e^y-e^{-y}}{2} = 2 \times \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2} \times \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}$

$\frac{e^x-e^{-x} + e^y-e^{-y}}{2} = \frac{(e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}})(e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}})}{2}$

$e^x-e^{-x} + e^y-e^{-y} = (e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}})(e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}})$

Can you go from here?

3. $
\frac{e^x-e^{-x}}{2} + \frac{e^y-e^{-y}}{2} =
2 \times \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2} \times \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}
" alt="
\frac{e^x-e^{-x}}{2} + \frac{e^y-e^{-y}}{2} = 2 \times \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2} \times \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}
" />

How did the 2 (highlighted in red) cancel out?

4. It cancels because of the multiplication of the denominators:
2/(2*2)= 1/2

Therefore, the two cancels.

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### sinhx sinhy=2sinh(x y/2)cosh(x-y/2)

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