# hyperbolic functions

• May 30th 2008, 04:46 AM
ArTiCK
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
• May 30th 2008, 05:11 AM
Jhevon
Quote:

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.
• May 30th 2008, 05:12 AM
bobak
Quote:

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
• May 30th 2008, 05:13 AM
mr fantastic
Quote:

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