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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).
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).
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(Original post by Mrepic Foulger)
Think of you identities relating these functions
Think of you identities relating these functions
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#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
Which identities do you mean? I've only just started learning this topic today so a little unsure on it
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#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
Which identities do you mean? I've only just started learning this topic today so a little unsure on it
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(Original post by Mrepic Foulger)
Cosh^2x - sinh^2x =1
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.
There are identities for the hyperbolic functions that are very similar to the ones for the circular (trigonometric) functions.
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) ?
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(Original post by Mrepic Foulger)
Cosh x can't be negative
Cosh x can't be negative
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#9
(Original post by beachpanda)
Ok, have I used the correct method then?
Ok, have I used the correct method then?
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(Original post by ThiagoBrigido)
What have you done for sinh(x) = 2 ?
What have you done for sinh(x) = 2 ?
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#11
(Original post by beachpanda)
My working is a couple of posts up
My working is a couple of posts up
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.
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#12
(Original post by beachpanda)
Ok, have I used the correct method then?
Ok, have I used the correct method then?
Last edited by mqb2766; 1 month ago
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(Original post by mqb2766)
Yes. When you square things up, you introduce extra an "solution" rather than lose anything.
Yes. When you square things up, you introduce extra an "solution" rather than lose anything.
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)
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#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)
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)
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.
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(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.
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.
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#17
(Original post by beachpanda)
My textbook says the answer is only +ve sqrt(5) ?
My textbook says the answer is only +ve sqrt(5) ?
"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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