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# Trigonometry equation [Solved!]

### 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

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

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

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

X

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

## Re: Trigonometry equation

X

Thanks for your help.