Question

hyperbolic trigonometric functions : "If sinhx= (15/8) and x>0, find the other hyperbolic functions at x"

CHECK ONLY...

Answer 1

\[\sinh{x}=\frac{15}{8},x\lt0\]\[\sinh{x}=\frac{e^{x}-e^{-x}}{2}=\frac{15}{8}...\]

Answer 2

check only? Do you have some work to show or something? :

Answer 3

am I going in the right direction?

Answer 4

Mmm.. yes this seems right.
Solve for x! :)

Answer 5

\[\frac{15}{8}=\frac{e^{x}-e^{-x}}{2}\rightarrow2(15)=8(e^{x}-e^{-x})\]does this work?

Answer 6

Sure, that's a good second step.
Let's divide a 2 out of each side,\[\large\rm 4e^x-4e^{-x}=15\]

Answer 7

The next step is a little tricky.
I won't spill the beans if you want to figure it out though :)

Any ideas? :)

Answer 8

\(4(e^{x}-e^{-x})=15\)?

Answer 9

Nooo silly :P
I distributed the 4, we don't want to move backwards hehe.

We have to apply some algebra trick to get rid of this negative exponent.

Answer 10

\[4e^{x}-\frac{4}{e^{x}}=15\]

Answer 11

So you rewrote it in another way, but he's still there... Hmm

Answer 12

Any way to rewrite the \(\large{e^{-x}}\) is what you mean?

Answer 13

Well I like how you wrote it in the denominator.
To get rid of denominator stuff, we multiply through by the value down there, ya? :)

Answer 14

So in this case, we'll multiply both sides of our equation by e^x.

Answer 15

Oh, I see what you are doing... removing any fractional forms for easier solving.

Answer 16

\[\large\rm e^x\left(4e^{x}-\frac{4}{e^{x}}\right)=15e^x\]

Answer 17

Ya it'll get rid of that fraction business which is great.

Answer 18

\[4e^{x}\times e^{x}=4e^{2x}\]yes?

Answer 19

and then treat it like a quadratic with \(x\) value as \(e^{x}\)

Answer 20

Yes, but it might actually be useful to keep the square on the outside.
\[\large\rm 4e^x\cdot e^x=4(e^x)^2\]If it's not immediately obvious why, that's ok.

Answer 21

Ahh ya there ya go :P too fast for me hehe

Answer 22

no, I get what you mean...
LOL sorry

Answer 23

\[\large\rm 4(e^x)^2-4=15e^x\]\[\large\rm 4(e^x)^2-15e^x-4=0\]So ya some sort of quadratic business going on.

Answer 24

Nonono don't do it this way!

When determining cosine (on the unit circle) when only given sine we would do the following:

\[\sin x = \frac{b}{c}\]
Use the pythagorean theorem to relate:
\[a^2+b^2=c^2\] so we would end up with:
\[\cos x = \frac{\sqrt{c^2-b^2}}{c}\]
then from here you get tangent from:
\[\tan x = \frac{\sin x}{\cos x}\]

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Similarly here we do the analogous thing, except instead of moving along a circle \(a^2+b^2=c^2\) you're moving along a hyperbola \(a^2-b^2=c^2\)!
\[\sinh x = \frac{b}{c}\]
Use the hyperbola to relate:
\[a^2-b^2=c^2\] so we would end up with:
\[\cosh x = \frac{\sqrt{c^2+b^2}}{c}\]
then from here you get tangent from:
\[\tanh x = \frac{\sinh x}{\cosh x}\]

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So you should check (using the exponentials, plug and chug 'em in) that you have:

\[\cosh^2 x -\sinh^2 t = 1\]

Since this is where this stands from.

Answer 25

No offense @zepdrix I like you haha, but doing it your way is like doing this:

\[\sin x = \frac{15}{8} \]\[\frac{e^{ix}-e^{-ix}}{2i}= \frac{15}{8} \]\[\cdots\]
in order to determine cosine, which feels wrong and is slower both for circles and the hyperbolic way too!

Answer 26

Ahhh crap good call :P

Literally all of the other hyperbolic identities slipped out of my brain when she mentioned the exponential one lol

Answer 27

We did a whole crap load of work to solve for x.
That doesn't really help us figure out cosh and tanh and those things.

Answer 28

I guess just for the fun of it, might as well just lay out some popular identities that are worth confirming at least once in everyone's life:

\[e^x =\cosh x + \sinh x\]\[1=\cosh^2 x - \sinh^2 x\]

Also, here's what I'm getting, your method should end up at the same answer once you solve the quadratic I think.

\[\cosh(x) = \frac{\sqrt{c^2+b^2}}{c} = \frac{17}{8}\]\[\tanh(x) = \frac{15}{17}\]

Answer 29

Hmm neato :d