how do you solve this simultaneous equations?

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

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- Nov 12th 2009, 01:14 AMalexandrabel90hyperbolic
how do you solve this simultaneous equations?

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12 - Nov 12th 2009, 02:24 AMcraig
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 AMalexandrabel90
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 AMmr fantastic