How Can I Prove This Hyperbolic Identity?

[thomas49th]

\equiv

Homework Statement

Hi, I've been given a hyperbolic identity to prove:

2sinhAsinhB \equiv Cosh(A+B) - Cosh(A-B)

Homework Equations

Cos(A\pm B) \equiv CosACosB \mp SinASinB

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 Cosh(A\pm B) \equiv CoshACoshB \mp SinhASinhB

Thanks :)

 

[tiny-tim]

Danger!

(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 )

 

[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 :)

 

[gabbagabbahey]

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

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

That is all you need to prove this identity...What are \sinh A and \sinh B? What does that make the LHS of your identity? What are \cosh(A+B) and \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.

 

[tiny-tim]

oops!

Hi 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!

 

[thomas49th]

using the identities i got

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

Thanks ;)

 

[gabbagabbahey]

using the identities i got

\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

 

[thomas49th]

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:
<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 :)

 

[tiny-tim]

2(\frac{e^{2x} - e^{-2x}}{2})(\frac{e^{2x} + e^{-2x}}{2})
…
<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? ​

 

[thomas49th]

but what about the original problem, with A and B? [/qoute]

hahah that's what I am asking you!
Not sure where to go now?
Any pointers :)

Thanks :)

 

[gabbagabbahey]

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

 

[thomas49th]

<br /> <br /> \sinh(B)=\frac{e^B-e^{-B}}{2}<br /> <br />

<br /> <br /> 2\sinh(A)\sinh(B)= \frac{1}{2} [ e^{AB} - 2e^{-AB} + e^{AB}]<br /> <br />

<br /> <br /> e^{AB} - e^{-AB}<br /> <br />

I'm almost there?
Thanks :)

 

[gabbagabbahey]

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

 

[thomas49th]

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) = e^{A+B}

 

[gabbagabbahey]

<br /> <br /> <br /> e^{A+B} - e^{A-B}<br /> <br /> <br />

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

 

[thomas49th]

arrrg it was just starting to look nice:

stick
-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!
:)

 

[gabbagabbahey]

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

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: \cosh(x)=\frac{e^x+e^{-x}}{2}...so \cosh (A+B)=___? And \cosh (A-B)=__? And so \cosh (A+B)-\cosh (A-B)=___?