# hyperbolic

• Nov 12th 2009, 01:14 AM
alexandrabel90
hyperbolic
how do you solve this simultaneous equations?

sinh x + sinh y = 25/12
cosh x - cosh y = 5/12
• Nov 12th 2009, 02:24 AM
craig
Quote:

Originally Posted by alexandrabel90
how do you solve this simultaneous equations?

sinh x + sinh y = 25/12
cosh x - cosh y = 5/12

You could try writing $\displaystyle \sin, \cos, \sinh, \cosh$ in their exponential form.

$\displaystyle \sinh{x} = \frac{e^{x}-e^{-x}}{2}$

$\displaystyle \cosh{x} = \frac{e^{x}+e^{-x}}{2}$

$\displaystyle \sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$

$\displaystyle \cos{x} = \frac{e^{ix}+e^{-ix}}{2}$

If you then multiply your second equation by 5 then you can set the two equal to one another.

I should just add that $\displaystyle i$ is a constant where $\displaystyle i = \sqrt{-1}$
• Nov 12th 2009, 02:42 AM
alexandrabel90
having using your mtd, i ended up with 2e^x + 2 e^-x = 3e^y -2e^y.

i tried to let e^x = a and e^y =b and then i got

2a + 2(1/a) = 3( b) - 2(1/b)

solving a as a quadratic eqn, i got a and b to be complex numbers..

which according to my notes, it is wrong as the answer should be real values.

did i do sth wrong?
• Nov 12th 2009, 02:48 AM
mr fantastic
Quote:

Originally Posted by alexandrabel90
how do you solve this simultaneous equations?

sinh x + sinh y = 25/12
cosh x - cosh y = 5/12

Solve the following simultaneously:

$\displaystyle a + b = \frac{25}{12}$ .... (1)

$\displaystyle c - d = \frac{5}{12}$ .... (2)

$\displaystyle c^2 - a^2 = 1$ .... (3)

$\displaystyle d^2 - b^2 = 1$ .... (4)