Hyperbolic question

By Anonymous on Tuesday, February 6, 2001 - 07:40 pm :

Hi,

Given that

y = \ln [\tan ((π / 4) + (x / 2))

show that $\sinh y = \tan x$ and $\cosh y = \sec x$ .
Thanks for your help.

By Kerwin Hui (Kwkh2) on Tuesday, February 6, 2001 - 08:12 pm :

Take exp() of both sides, and work out also its reciprocal. Then plug into the definition of sinh and cosh to see it works out nicely.

Kerwin

By Anonymous on Tuesday, February 6, 2001 - 09:31 pm :

I had a go, but got stuck.

e^{y} = \tan ((π / 4) + (x / 2))

$\sinh y = (e^{y} - e^{- y}) / 2$
I also put in the tan part into the y, but was not sure how to show that this equalled tan x?

By Michael Doré (Md285) on Wednesday, February 7, 2001 - 01:33 pm :

Hi,

The main formula you need to know for this question is the addition angle formula for $\tan x$ :

$\tan (a + b) = (\tan a + \tan b) / (1 - \tan a \tan b)$

Of course we also need the definition of cosh and sinh:

$2 \cosh x = e^{x} + e^{- x}$

$2 \sinh x = e^{x} - e^{- x}$

Now let's start off by looking at $\tan (π / 4 + x / 2)$ :

$\tan (π / 4 + x / 2) = (\tan π / 4 + \tan x / 2) / (1 - \tan π / 4 \tan x / 2) = (1 + \tan x / 2) / (1 - \tan x / 2)$

using the addition angle formula for tan.

$e^{y} = (1 + \tan x / 2) / (1 - \tan x / 2)$ [1]

$e^{- y} = (1 - \tan x / 2) / (1 + \tan x / 2)$ [2]

Consider [1] + [2]:

$2 \cosh y = [(1 + \tan x / 2)^{2} + (1 - \tan x / 2)^{2}] / (1 - \tan^{2} x / 2)$

(by finding a common denominator)

So:

$2 \cosh y = [2 + 2 \tan^{2} x / 2] / (1 - \tan^{2} x / 2) = (2 \sec^{2} x / 2) / (1 - \tan^{2} x / 2)$

$= (2 \tan x / 2) / (1 - \tan^{2} x / 2) \times (\sec^{2} x / 2) / (\tan x / 2)$

$= \tan x / (\cos x / 2 \sin x / 2)$

$= \tan x / (1 / 2 \sin x)$ (using $\sin x = 2 \sin x / 2 \cos x / 2$ )

$= 2 \sec x$

As required.

Now consider [1] - [2]:

$2 \sinh y = [(1 + \tan x / 2)^{2} - (1 - \tan x / 2)^{2}] / (1 - \tan^{2} x / 2) = (4 \tan x / 2) / (1 - \tan^{2} x / 2) = 2 \tan x$

Hope this helps,

Michael D.

By Anonymous on Wednesday, February 7, 2001 - 05:31 pm :

Thank you Michael.
I am going to read you post in detail.
I'll get back if I get any probs. Is that ok?

By Michael Doré (Md285) on Wednesday, February 7, 2001 - 11:19 pm:

Of course :-) And I'm certain there should be a more elegant way of doing it than the way I did it, so if you or anyone else have any further ideas, please post them!

Yours,

Michael