Hyperbolic Equations Question - Further

Watch
beachpanda
Badges: 12
Rep:
?
#1
Report Thread starter 1 month ago
#1
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).
0
reply
Mrepic Foulger
Badges: 17
Rep:
?
#2
Report 1 month ago
#2
Think of you identities relating these functions
0
reply
beachpanda
Badges: 12
Rep:
?
#3
Report Thread starter 1 month ago
#3
(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
0
reply
Mrepic Foulger
Badges: 17
Rep:
?
#4
Report 1 month ago
#4
(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
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#5
Report 1 month ago
#5
(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.
0
reply
beachpanda
Badges: 12
Rep:
?
#6
Report Thread starter 1 month ago
#6
(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) ?
0
reply
Mrepic Foulger
Badges: 17
Rep:
?
#7
Report 1 month ago
#7
Cosh x can't be negative
0
reply
beachpanda
Badges: 12
Rep:
?
#8
Report Thread starter 1 month ago
#8
(Original post by Mrepic Foulger)
Cosh x can't be negative
Ok, have I used the correct method then?
0
reply
ThiagoBrigido
Badges: 16
Rep:
?
#9
Report 1 month ago
#9
(Original post by beachpanda)
Ok, have I used the correct method then?
What have you done for sinh(x) = 2 ?
0
reply
beachpanda
Badges: 12
Rep:
?
#10
Report Thread starter 1 month ago
#10
(Original post by ThiagoBrigido)
What have you done for sinh(x) = 2 ?
My working is a couple of posts up
0
reply
ThiagoBrigido
Badges: 16
Rep:
?
#11
Report 1 month ago
#11
(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.
0
reply
mqb2766
Badges: 19
Rep:
?
#12
Report 1 month ago
#12
(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.
Last edited by mqb2766; 1 month ago
0
reply
beachpanda
Badges: 12
Rep:
?
#13
Report Thread starter 1 month ago
#13
(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)
0
reply
mqb2766
Badges: 19
Rep:
?
#14
Report 1 month ago
#14
(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.
0
reply
Nkelly103
Badges: 6
Rep:
?
#15
Report 1 month ago
#15
Name:  image.jpg
Views: 4
Size:  139.2 KB This is a solution using the definitions of Sinhx and coshx.
1
reply
beachpanda
Badges: 12
Rep:
?
#16
Report Thread starter 1 month ago
#16
(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!

(Original post by Nkelly103)
Name:  image.jpg
Views: 4
Size:  139.2 KB This is a solution using the definitions of Sinhx and coshx.
That's v handy, thankyou!
0
reply
DFranklin
Badges: 18
Rep:
?
#17
Report 1 month ago
#17
(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].
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (209)
57.26%
I don't have everything I need (156)
42.74%

Watched Threads

View All
Latest
My Feed