Home  |  About  | Last |  Submit  |  Contact
AllQuests.com



Previous Question:  Windows Remaking Paging File  Web Hosting LoungeNext Question:  Favorite website to relax at  Web Hosting Lounge
Question Proving hyperbolic identity ( Physics Forums Calculus Beyond )
Updated: 2010-07-25 03:50:22 (16)
Proving hyperbolic identity

\equiv1. The problem statement, all variables and given/known data
Hi, I've been given a hyperbolic identity to prove:

LaTeX Code: 2sinhAsinhB \\equiv Cosh(A+B) - Cosh(A-B)


2. Relevant equations

LaTeX Code: Cos(A\\pm B) \\equiv CosACosB \\mp SinASinB

3. The attempt at a solution

I have Cosh(A+B) and - Cosh(A-B) so i can kind of see that there will be two lots of SinhASinhB and the CoshACoshB will cancel, but how do I prove it? I mean how do I know that LaTeX Code: Cosh(A\\pm B) \\equiv CoshACoshB \\mp SinhASinhB

Thanks :)

Answers: Proving hyperbolic identity ( Physics Forums Calculus Beyond )
Proving hyperbolic identity

(have a ? )

Warning, warning, thomas49th!

Cosh(A?B) = coshAcoshB ? sinhAsinhB (the opposite sign to cos).

(this is because, from Euler's formula cosx = coshix, isinx = sinhix, so i2sinAsinB = sinhAsinhB )

tiny-tim

Proving hyperbolic identity

Errr sorry I don't follow :S. I've only just started hyperbolics today and havn't used an imaginary numbers with them yet?

Thanks :)

thomas49th

Proving hyperbolic identity

I assume that you have been taught the definitions of these functions:

LaTeX Code: \\cosh(x)=\\frac{e^x+e^{-x}}{2} and LaTeX Code: \\sinh(x)=\\frac{e^x-e^{-x}}{2}

That is all you need to prove this identity....What are LaTeX Code: \\sinh  A and LaTeX Code: \\sinh B ? What does that make the LHS of your identity? What are LaTeX Code: \\cosh(A+B) and LaTeX Code: \\cosh(A-B) ? What does that make the RHS of your identity? Can you show that the two expressions are equivalent? If so, then you prove the identity.

gabbagabbahey

Proving hyperbolic identity

Hi thomas49th!
Originally Posted by thomas49th
Errr sorry I don't follow :S. I've only just started hyperbolics today and havn't used an imaginary numbers with them yet?

Thanks :)
Oops!

In that case, all you need to know is that "hyperbolic" trig functions cosh sinh tanh sech coth and cosech work almost the same as ordinary trig functions (for example, sinh(2x) = 2sinhx coshx), but occasionally you get a + instead of a minus (or vice versa) I think only when you have two sinh's.

But, to be on the safe side, use gabbagabbahey's method!

tiny-tim

Proving hyperbolic identity

using the identities i got

LaTeX Code: \\frac{1}{2} e^{2x} - e^{-2x}
which is equilivlent to cosh2x. but where next?

Thanks ;)

thomas49th

Proving hyperbolic identity

Originally Posted by thomas49th
using the identities i got

LaTeX Code: \\frac{1}{2} e^{2x} - e^{-2x}
which is equilivlent to cosh2x. but where next?

Thanks ;)
I think you'd better show me your work

gabbagabbahey

Proving hyperbolic identity

LaTeX Code: 2(\\frac{e^{2x} - e^{-2x}}{2})(\\frac{e^{2x} + e^{-2x}}{2})
one 2 cancels so you get a half overall. the difference of two squares acts nicely leaving us with:
LaTeX Code: <BR>      \\frac{1}{2} e^{2x} - e^{-2x}<BR>
and i was looking back over the notes in class and I saw that we identified cosh2x as that

Right?
Thanks :)

thomas49th

Proving hyperbolic identity

Originally Posted by thomas49th
LaTeX Code: 2(\\frac{e^{2x} - e^{-2x}}{2})(\\frac{e^{2x} + e^{-2x}}{2})

LaTeX Code: <BR>      \\frac{1}{2} e^{2x} - e^{-2x}<BR>
and i was looking back over the notes in class and I saw that we identified cosh2x as that

Right?
Thanks :)
oh i see you're proving 2 sinhx coshx = sinh 2x (not cosh 2x! cosh is the positive one )
but what about the original problem, with A and B?

tiny-tim

Proving hyperbolic identity

[quote=thomas49th;1975600]
but what about the original problem, with A and B? [/qoute]
hahah that's what im asking you!
Not sure where to go now?
Any pointers :)

Thanks :)

thomas49th

Proving hyperbolic identity

Well, if LaTeX Code: <BR>      \\sinh(x)=\\frac{e^x-e^{-x}}{2}<BR>       ....then LaTeX Code: <BR>      \\sinh(A)=\\frac{e^A-e^{-A}}{2}<BR>       ....so LaTeX Code: \\sinh(B)= ___? And so LaTeX Code: <BR>      2\\sinh(A)\\sinh(B)= ___?

gabbagabbahey

Proving hyperbolic identity

LaTeX Code: <BR><BR>            \\sinh(B)=\\frac{e^B-e^{-B}}{2}<BR>            <BR>

LaTeX Code: <BR><BR>            2\\sinh(A)\\sinh(B)=   \\frac{1}{2} [ e^{AB} - 2e^{-AB} + e^{AB}]<BR>            <BR>

LaTeX Code: <BR><BR>            e^{AB} - e^{-AB}<BR>            <BR>

I'm almost there?
Thanks :)

thomas49th

Proving hyperbolic identity

Hmmm... LaTeX Code: e^Ae^B=e^{A+B} \\neq e^{AB}

gabbagabbahey

Proving hyperbolic identity

LaTeX Code: <BR><BR><BR>                  e^{A+B} - e^{A-B}<BR>                  <BR>            <BR>

Notice how there is an A+B and A-B from JUST like in cos(A+B)

now how do I show that cos(A+B) = LaTeX Code: e^{A+B}

thomas49th

Proving hyperbolic identity

Originally Posted by thomas49th
LaTeX Code: <BR><BR><BR>                  e^{A+B} - e^{A-B}<BR>                  <BR>            <BR>

aren't you missing a couple of terms in that expression?

gabbagabbahey

Proving hyperbolic identity

arrrg it was just starting to look nice:

stick
LaTeX Code:  -e^{-A+B} -e^{-A-B}

I've goto go to bed... knackard sorry. it's almost midnight ere in merry old england

i'd read any other message people post on here in the morning. thanks for everything!
:)

thomas49th

Proving hyperbolic identity

Looks good, except your missing a factor of 1/2, and one of your signs is incorrect..... you should have:

LaTeX Code: 2\\sinh (A) \\sinh (B)=\\frac{e^{A+B}-e^{A-B}-e^{-A+B}+e^{-A-B}}{2}

You also know the definition of cosh: LaTeX Code: \\cosh(x)=\\frac{e^x+e^{-x}}{2} ....so LaTeX Code: \\cosh (A+B)= ___? And LaTeX Code: \\cosh (A-B)= __? And so LaTeX Code: \\cosh (A+B)-\\cosh (A-B)= ___?

gabbagabbahey

Previous Question:  Windows Remaking Paging File  Web Hosting Talk  Web Hosting LoungeNext Question:  Favorite website to relax at  Web Hosting Talk  Web Hosting Lounge

- Source: Proving hyperbolic identity Physics Forums Calculus Beyond
- Previous Question: Windows Remaking Paging File Web Hosting Talk Web Hosting Lounge
- Next Question: Favorite website to relax at Web Hosting Talk Web Hosting Lounge