# Trigonometry equation [Solved!]

**Lee** 25 Nov 2015, 09:38

### My question

can you help with following equation

simplify

1+sinh2A+cosh2A / 1-sinh2A-cosh2A

### Relevant page

5. Trigonometric Equations

### What I've done so far

Looked for similar examples in IntMath, but couldn't find any.

X

can you help with following equation
simplify
1+sinh2A+cosh2A / 1-sinh2A-cosh2A

Relevant page
<a href="/analytic-trigonometry/5-trigonometric-equations.php">5. Trigonometric Equations</a>
What I've done so far
Looked for similar examples in IntMath, but couldn't find any.

## Re: Trigonometry equation

**Murray** 26 Nov 2015, 04:02

Hello Lee

Your expression is actually related to Exponential Functions rather than trigonometric ones. I don't include the topic of Hyperbolic Functions on my site yet, but you may find the following helpful:

Hyperbolic function (Wikipedia)

BTW, you are missing brackets in your question. The brackets make a huge difference to the meaning of the algebra. I think you meant this:

(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)

Using the math input system, it would look like this:

`(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)`

Without brackets, you get this:

`1+sinh2A+cosh2A / 1-sinh2A-cosh2A `

Anyway, to get you started, you need to know that

`sinh x = (e^x-e^-x)/2`

and

`cosh x = (e^x+e^-x)/2`

So the top line of your fraction will be:

`1+sinh2A+cosh2A = 1 + (e^(2A)-e^(-2A))/2 + (e^(2A)+e^(-2A))/2`

Simplify this, then do a similar thing for the bottom of the fraction.

Then see what you are left with, after simplifying the whole thing.

I hope that makes sense.

Regards

X

Hello Lee
Your expression is actually related to Exponential Functions rather than trigonometric ones. I don't include the topic of Hyperbolic Functions on my site yet, but you may find the following helpful:
<a href="https://en.wikipedia.org/wiki/Hyperbolictrigonometricfunction">Hyperbolic function (Wikipedia)</a>
BTW, you are missing brackets in your question. The brackets make a huge difference to the meaning of the algebra. I think you meant this:
(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)
Using the math input system, it would look like this:
`(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)`
Without brackets, you get this:
`1+sinh2A+cosh2A / 1-sinh2A-cosh2A `
Anyway, to get you started, you need to know that
`sinh x = (e^x-e^-x)/2`
and
`cosh x = (e^x+e^-x)/2`
So the top line of your fraction will be:
`1+sinh2A+cosh2A = 1 + (e^(2A)-e^(-2A))/2 + (e^(2A)+e^(-2A))/2`
Simplify this, then do a similar thing for the bottom of the fraction.
Then see what you are left with, after simplifying the whole thing.
I hope that makes sense.
Regards

## Re: Trigonometry equation

**Lee** 27 Nov 2015, 01:52

OK, thanks.

`1 + (e^(2A)-e^(-2A))/2 + (e^(2A)+e^(-2A))/2`

`= 1 + e^(2A)`

The bottom of the fraction is:

`1-sinh2A-cosh2A `

`= 1 - ((e^(2A)-e^(-2A))/2) - ((e^(2A)+e^(-2A))/2)`

`= 1 - e^(2A)`

So overall, it's

`(1 + e^(2A))/(1 - e^(2A))`

Is that the end?

X

OK, thanks.
`1 + (e^(2A)-e^(-2A))/2 + (e^(2A)+e^(-2A))/2`
`= 1 + e^(2A)`
The bottom of the fraction is:
`1-sinh2A-cosh2A `
`= 1 - ((e^(2A)-e^(-2A))/2) - ((e^(2A)+e^(-2A))/2)`
`= 1 - e^(2A)`
So overall, it's
`(1 + e^(2A))/(1 - e^(2A))`
Is that the end?

## Re: Trigonometry equation

**Murray** 27 Nov 2015, 13:33

Yes, that's as far as you can simplify it.

X

Yes, that's as far as you can simplify it.

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