Trigonometry equation [Solved!]

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.

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

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?

X

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

X

Thanks for your help.