1. ## hyperbolic functions

Hi all,

I am stuck with this question:

Using coshx and sinhx to verify the identity:

(coshx + sinhx)^n = cosh(nx) + sinh(nx)

I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

ArTiCK

2. Originally Posted by ArTiCK
Hi all,

I am stuck with this question:

Using coshx and sinhx to verify the identity:

(coshx + sinhx)^n = cosh(nx) + sinh(nx)

I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

ArTiCK
plug in the identities you have into the right side first and simplify as much as possible. then do the same for the left side, and show that they simplify to the same thing.

3. Originally Posted by ArTiCK
By writing $\cosh x$ and $\sinh x$ in exponential form prove the identity:

$(\cosh x + \sinh x)^n = \cosh nx + \sinh nx
$
$\cosh x = \frac{1}{2}(e^{x} + e^{-x})$ and $\sinh x = \frac{1}{2}(e^{x} - e^{-x})$

$\therefore coshx + sinhx = \frac{1}{2}(e^{x} + e^{-x}) + \frac{1}{2}(e^{x} - e^{-x})$
$\Rightarrow e^x$

$(coshx + sinhx)^n = e^{nx}$

$\Rightarrow \frac{1}{2}e^{nx} + \frac{1}{2}e^{nx} + \underbrace{\frac{1}{2}e^{-nx} -\frac{1}{2}e^{-nx}}$
. . . . . . . . . . . adding "zero"

$\Rightarrow \frac{1}{2}(e^{nx} +e^{-nx}) + \frac{1}{2}(e^{nx} -e^{-nx})$

$\Rightarrow \cosh nx + \sinh nx$

Bobak

4. Originally Posted by ArTiCK
Hi all,

I am stuck with this question:

Using coshx and sinhx to verify the identity:

(coshx + sinhx)^n = cosh(nx) + sinh(nx)

I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

Note: $\cosh (ax) + \sinh (ax) = \frac{e^{ax} + e^{-ax}}{2} + \frac{e^{ax} - e^{-ax}}{2} = e^{ax}$.