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Hyperbolic Equations Question - Further

Can anyone help me with the below question?

I tried solving for x in sinh(x) = 2 first, then subbing my solution into cosh(x) but couldn't get it.

Is that the correct method? Or is there a different way to do it?

Given that sinh(x)=2, find the exact values of cosh(x) and tan(x).
Think of you identities relating these functions
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Avatar for beachpanda
beachpanda
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Original post by Mrepic Foulger
Think of you identities relating these functions


Which identities do you mean? I've only just started learning this topic today so a little unsure on it
Original post by beachpanda
Which identities do you mean? I've only just started learning this topic today so a little unsure on it


Cosh^2x - sinh^2x =1
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davros
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Original post by beachpanda
Which identities do you mean? I've only just started learning this topic today so a little unsure on it

There are identities for the hyperbolic functions that are very similar to the ones for the circular (trigonometric) functions.
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beachpanda
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Original post by Mrepic Foulger
Cosh^2x - sinh^2x =1


Original post by davros
There are identities for the hyperbolic functions that are very similar to the ones for the circular (trigonometric) functions.


Ah yeah I have seen that - I've used the identity to get this:

sinh(x) = 2
sinh^2(x) = 4

cosh^2(x) - sinh^2(x) = 1
cosh^2(x) = 1 + sinh^2(x)
cosh^2(x) = 1 + 4
cosh^2(x) = 5
cosh(x) = +/- sqrt(5)

My textbook says the answer is only +ve sqrt(5) ?
Cosh x can't be negative
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beachpanda
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Original post by Mrepic Foulger
Cosh x can't be negative


Ok, have I used the correct method then?
Original post by beachpanda
Ok, have I used the correct method then?

What have you done for sinh(x) = 2 ?
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beachpanda
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Original post by ThiagoBrigido
What have you done for sinh(x) = 2 ?


My working is a couple of posts up
Original post by beachpanda
My working is a couple of posts up

It seems to be that by squaring both sides sinhx=2, you have lost some information along the way.
Use the fact that sinh(x) = 2 => (e^x - e^-x)/2 = 2. You should notice that by rearranging the equation you will get a quadratic equation disguised in e^x. By knowing that ln is the inverse of e, you can multiply both sides by ln, which gives you the value for x = ln(2+-√5) ; ln(-√5) is undefined you can discard it, therefore sub the value of x into coshx, which by the way is (e^x + e^-x) / 2, you can enter that on your calculator which will give you 2.23606...., which in fact is √5 . Do the same for tanhx. Hope that helps.
Original post by beachpanda
Ok, have I used the correct method then?

Yes. When you square things up, you introduce an extra "solution" rather than lose anything.
(edited 3 years ago)
Original post by mqb2766
Yes. When you square things up, you introduce extra an "solution" rather than lose anything.

I see, because the graph of cosh(x) is always above the x-axis it can't have a negative output, so -sqrt(5) is invalid?

Does this method look right for tanh(x)? My textbook says the answer is tanh(x) = 2 / sqrt(5)

tanh(x) = sinh(x) / cosh(x)
sinh(x) = 2
cosh(x) = sqrt(5)
tanh(x) = 2 / sqrt(5)
Original post by beachpanda
I see, because the graph of cosh(x) is always above the x-axis it can't have a negative output, so -sqrt(5) is invalid?

Does this method look right for tanh(x)? My textbook says the answer is tanh(x) = 2 / sqrt(5)

tanh(x) = sinh(x) / cosh(x)
sinh(x) = 2
cosh(x) = sqrt(5)
tanh(x) = 2 / sqrt(5)

Sure, hard to go wrong with that tanh calculation?

It's not surprising that you ignore some solutions which come out of quadratic identities. If you were given cosh and had to determine sinh, you'd have to consider both +/- solutions. Just learn the basic curves and it will become familiar.
image.jpg This is a solution using the definitions of Sinhx and coshx.
Original post by mqb2766
Sure, hard to go wrong with that tanh calculation?

It's not surprising that you ignore some solutions which come out of quadratic identities. If you were given cosh and had to determine sinh, you'd have to consider both +/- solutions. Just learn the basic curves and it will become familiar.

Nice one, thanks again! :biggrin:

Original post by Nkelly103
image.jpg This is a solution using the definitions of Sinhx and coshx.

That's v handy, thankyou!
Original post by beachpanda
My textbook says the answer is only +ve sqrt(5) ?

To add to the other comments, note that if you had the similar trig question:

"sin(x) = .6, what is cos(x)?", you'd have a similar ambiguity about whether cos(x) was 0.8 or -0.8.

The only real difference with hyperbolics is that you know cosh(x) >= 1, so you can rule out the -ve answer.

But if you had instead "cosh(x) = sqrt(5), what is sinh(x)?", both 2 and -2 would be possible answers.

Also, although you *can* reduce to e^x, or even solve for x and then just evaluate the function, if you have a question with hyperbolic functions where replacing "sinh/cosh/tanh by sin/cos/tan" would leave you with something that looks like it should be solved with trig identities, then you are *probably* expected to use the hyperbolic analogues to the trig identities.

[c.f. you *could* calculate cos(arcsin(0.6)) in my hypothetical trig example, but it wouldn't be the expected solution].

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