# Hyperbolic

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#1
Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Hence find the real values of x and y (x=ln 3, y=ln 2)
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1 month ago
#2
Try using the define of sinh and cosh to rewrite all the terms. Then use a method similar to how you would solve simultaneous equations to get your two desired equations
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#3
I know all of that. Second step (simultaneous equation is a joke) But first step (transformation) is pretty hard. I have got at least 100 "help" like that. Can you be more specific? Or simply say I do not know!
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1 month ago
#4
(Original post by Hillsby)
I know all of that. Second step (simultaneous equation is a joke) But first step (transformation) is pretty hard. I have got at least 100 "help" like that. Can you be more specific? Or simply say I do not know!
Well if you know all of that, then why are you struggling
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#5
I am struggling with this part:

Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Can you solve simultaneous equation are easy. I meant that basic equations like sinh =(e^x-e^-x)/2 are not helpful, but full working. You can try identities and multiplication by e^x. Q is the part of the most difficult Qs of A-level and uni. If you ask, they usually turn to quadratic hyperbolic or double angle hyperbolic.Anyway, thank you for interest
0
1 month ago
#6
(Original post by Hillsby)
I am struggling with this part:

Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Can you solve simultaneous equation are easy. I meant that basic equations like sinh =(e^x-e^-x)/2 are not helpful, but full working. You can try identities and multiplication by e^x. Q is the part of the most difficult Qs of A-level and uni. If you ask, they usually turn to quadratic hyperbolic or double angle hyperbolic.Anyway, thank you for interest
You don't need to manipulate each expression past turning sinh and cosh into their respective exponential counterparts. All you have to do afterwards is add/subtract the 2 equations to get the desired result. Why would you change the expressions anymore than that if the answer they want is of the form e^x, e^-x, e^y, e^-y? If that's not what you're asking for help with then please be more clear.
1
#7
(Original post by JustSomeGuy:/)
You don't need to manipulate each expression past turning sinh and cosh into their respective exponential counterparts. All you have to do afterwards is add/subtract the 2 equations to get the desired result. Why would you change the expressions anymore than that if the answer they want is of the form e^x, e^-x, e^y, e^-y? If that's not what you're asking for help with then please be more clear.
I copied full question for you: 1. You have to prove as showed
2. you have to find x, y.
to find only x, y is marked 50% or E (see bellow)

Is it clear now?

Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Hence find x and y
0
1 month ago
#8
(Original post by Hillsby)
I copied full question for you: 1. You have to prove as showed
2. you have to find x, y.
to find only x, y is marked 50% or E (see bellow)

Is it clear now?

Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Hence find x and y
Have you even attempted the problem?
0
1 month ago
#9
(Original post by Hillsby)
I copied full question for you: 1. You have to prove as showed
2. you have to find x, y.
to find only x, y is marked 50% or E (see bellow)

Is it clear now?

Given that:

sinh x + sinh y = 25/12

cosh x - cosh y = 5/12

show that:

2e^x=5+2e^-y

and

3e^-x=5+3e^y

Hence find x and y
The question was already clear. What I meant was it's not clear what part you're struggling with.
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#10
(Original post by JustSomeGuy:/)
The question was already clear. What I meant was it's not clear what part you're struggling with.
So, do you think you ca solve it

(Original post by tej3141)
Have you even attempted the problem?
O ye, cannot solve it (first part). Find x and y is not problem
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1 month ago
#11
(Original post by Hillsby)
So, do you think you ca solve it

O ye, cannot solve it (first part). Find x and y is not problem
I already tried and solved it beforehand. As I said before, all you do is convert both equations into their exponential forms (nothing more) and then you add the 2 equations together for one of the answers and subtract them for the other (as I had already stated in the previous post).
1
1 month ago
#12
(Original post by Hillsby)
So, do you think you ca solve it

O ye, cannot solve it (first part). Find x and y is not problem
How did you find x and y without answering answering first part?
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1 month ago
#13
(Original post by tej3141)
How did you find x and y without answering answering first part?
Most likely used the equations that he was supposed to prove existed from part a.
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1 month ago
#14
(Original post by JustSomeGuy:/)
Most likely used the equations that he was supposed to prove existed from part a.
Yh I know. I didn't mean it like that. I meant the 2 parts seem to be of the same complexity to me
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#15
(Original post by JustSomeGuy:/)
I already tried and solved it beforehand. As I said before, all you do is convert both equations into their exponential forms (nothing more) and then you add the 2 equations together for one of the answers and subtract them for the other (as I had already stated in the previous post).
Show me your working! Convert both equations into their exponential form, stated in the task.
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1 month ago
#16
(Original post by Hillsby)
Show me your working! Convert both equations into their exponential form, stated in the task.
Perform 1+2 for one equation and 1-2 for the other. What about it do you find difficult? You're just using the properties of sinh and cosh and then doing some simultaneous equations which you said was easy in your first post. Anywhere hope this clears things up...

Last edited by JustSomeGuy:/; 1 month ago
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#17
(Original post by JustSomeGuy:/)
Perform 1+2 for one equation and 1-2 for the other. What about it do you find difficult? You're just using the properties of sinh and cosh and then doing some simultaneous equations which you said was easy in your first post. Anywhere hope this clears things up...

A good start, but what next? Remember, you have to get second pair of equation exactly it is given. Good luck! you can use substitution anytime if you are not sure.
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1 month ago
#18
(Original post by Hillsby)
A good start, but what next? Remember, you have to get second pair of equation exactly it is given. Good luck! you can use substitution anytime if you are not sure.
They've already told you exactly how to do it multiple times - are you still stuck? People aren't meant to share full solutions in the study help forums by the way, hence why they're giving you hints instead of just giving you the answer
Last edited by Interea; 1 month ago
1
1 month ago
#19
Yeah I'm quite confused why the one who's asking for help is trying to give hints here
1
#20
(Original post by Interea)
They've already told you exactly how to do it multiple times - are you still stuck? People aren't meant to share full solutions in the study help forums by the way, hence why they're giving you hints instead of just giving you the answer
Did they??I did not notice. Can you copy it for me..
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