hyperbolic

**alexandrabel90** · November 12th 2009, 01:14 AM

how do you solve this simultaneous equations?

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

**craig** · November 12th 2009, 02:24 AM

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 $\sin, \cos, \sinh, \cosh$ in their exponential form.

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

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

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

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

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

I should just add that $i$ is a constant where $i = \sqrt{-1}$

**alexandrabel90** · November 12th 2009, 02:42 AM

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?

**mr fantastic** · November 12th 2009, 02:48 AM

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:

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

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

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

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

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