# Noooo, I got wasted o.O

• Nov 23rd 2008, 02:24 AM
Lonehwolf
Noooo, I got wasted o.O
Prove teh Identitiy

$
cosh (x+ y) = cosh x cosh y + sinh x sinh y
$

from the basic definitions of the hyperbolic functions. Hence, or otherwise show that if x and y satisfy the equations

cosh x cosh y = 2,
cosh x sinh y = -1

then x = -y = +or- ln (1+*squareroot* of 2)

I dunno how to use the math thingy too well so xD sorry if it's messed up.

This question just has me staring, don't even know what I should be doing.
• Nov 23rd 2008, 02:37 AM
Moo
Hello,
Quote:

Originally Posted by Lonehwolf
Prove teh Identitiy

$
cosh (x+ y) = cosh x cosh y + sinh x sinh y
$

from the basic definitions of the hyperbolic functions.

Use the "basic definitions" :
$\cosh(x)=\frac{e^x+e^{-x}}{2}$
$\sinh(x)=\frac{e^x-e^{-x}}{2}$

Start from the right-hand side of the equality if it can help you.

~~~~~~~~~~~~~~~~~~~~
also remember these rules :
$a^b a^c=a^{b+c}$
$a^0=1$
• Nov 23rd 2008, 02:44 AM
Moo
Quote:

Hence, or otherwise show that if x and y satisfy the equations

cosh x cosh y = 2,
cosh x sinh y = -1

then x = -y = +or- ln (1+*squareroot* of 2)

I dunno how to use the math thingy too well so xD sorry if it's messed up.

This question just has me staring, don't even know what I should be doing.
shouldn't it be sinh(x)sinh(y)=-1 for the second equation ?

If so, just add the two equation :
$\cosh(x)\cosh(y)+\sinh(x)\sinh(y)=1$

But the left-hand side of the equation is just $\cosh(x+y)$

So $1=\cosh(x+y)=\frac{e^{x+y}+e^{-(x+y)}}{2}$

Let $X=e^{x+y}$ :

$1=\frac{X+\frac 1X}{2}$
multiply both sides by $2X$ :
$2X=X^2+1$

$X^2-2X+1=(X-1)^2=0$

hence $X=e^{x+y}=1$

so $\boxed{x+y=0}$

__________________________________
You can see that $\cosh(-x)=\cosh(x)$

Hence $2= \cosh(x)\cosh(y)=\cosh(x)\cosh(-x)=\left(\cosh(x)\right)^2$

So $\cosh(x)=\pm \sqrt{2}$

Can you try to do this one ? There's just a quadratic to solve :)
• Nov 27th 2008, 10:37 PM
Lonehwolf
Sorry for the late reply, ended up making a format and was left for three days without any access to internet >.<

Either way, I'll use the first question's guidance you gave me to it, hope it ends up somewhere xD

For the second one, yeh, it was sinh(x)sinh(y) = -1

Pretty good of you to notice my... uhh... testing the community? (Wondering)

But since I have a private lesson now I can't take any time trying those, it was from an Advanced Level past paper, and golly gosh you're good >_<
• Nov 28th 2008, 08:12 AM
Lonehwolf
Ok, got the time to check them out and I solved both successfully :D

Thanks moo, tyvm :)