help with this domain

**Raoh** · June 20th 2009, 12:41 PM

hi

i'm trying to find the domain of this function :
let $\mathbb{E} = [-\pi,+\pi]$
$x \in \mathbb{E} ;$ $f(x) = \sqrt{\frac{1+\cos \left ( x \right )}{1-\cos \left ( x \right )}}$
i started by this :
$D = \left \{ x\in \mathbb{E}\mid 1-cos(x)\neq 0 , \frac{1+cos(x)}{1-cos(x)}\geq 0 \right \}$

which means $1-cos(x) > 0$ and $1+cos(x) \geq 0$ because we have $-1\leq cos(x)\leq 1$
what should i do next ?
thanks a lot

**running-gag** · June 20th 2009, 12:59 PM

Originally Posted by Raoh

hi

i'm trying to find the domain of this function :
let $\mathbb{E} = [-\pi,+\pi]$
$x \in \mathbb{E} ;$ $f(x) = \sqrt{\frac{1+\cos \left ( x \right )}{1-\cos \left ( x \right )}}$
i started by this :
$D = \left \{ x\in \mathbb{E}\mid 1-cos(x)\neq 0 , \frac{1+cos(x)}{1-cos(x)}\geq 0 \right \}$

which means $1-cos(x) > 0$ and $1+cos(x) \geq 0$ because we have $-1\leq cos(x)\leq 1$
what should i do next ?
thanks a lot

Hi

$1+cos(x) \geq 0$ is always true

$1-cos(x) > 0$ is true unless cos x = 1 which means x = 0

Therefore $D = [-\pi,+\pi]$ \ {0}

Note that you could use the formula $1+cos(x) = 2 \cos^2 \frac{x}{2}$ and $1-cos(x) = 2 \sin^2 \frac{x}{2}$

Then $f(x) = \sqrt{\frac{1+\cos \left ( x \right )}{1-\cos \left ( x \right )}} = \frac{\left|\cos \frac{x}{2}\right|}{\left|\sin \frac{x}{2}\right|}$

**Plato** · June 20th 2009, 01:02 PM

This should tell you a great deal.
$- 1 \leqslant - \cos (x) \leqslant 1\; \Rightarrow \;0 \leqslant 1 - \cos (x) \leqslant 2$

**VonNemo19** · June 20th 2009, 01:43 PM

Originally Posted by Raoh

hi

i started by this :
$D = \left \{ x\in \mathbb{E}\mid 1-cos(x)\neq 0 , \frac{1+cos(x)}{1-cos(x)}\geq 0 \right \}$

If you are able to use such impressive vernacular, as stated in this qoute, then I am supremely confident that you can find all x for which the function is defined.

But, here's my 2 cents anyway

We must find all values of x such that the quotient inside the radical is greater than zero, while the denominator is not equal to zero.

A quotient is positve only if the numerator and denomiator are both positive, or both negative.

$1+\cos{x}$ is always greator to or equal to zero. This fact makes are task much easier. Now, all we have to do is find when $1-\cos{x}\leq0$

We Know that the only place at which 1-cos(x) is zero in the interval $-\pi{\leq}x\leq\pi$ is at $x=0$ , and greator than zero every where else.

eg $1-\cos({-\pi})=1-(-1)=2$

So the domain of the function is $[-\pi,0)U(0,\pi]$

**Plato** · June 20th 2009, 01:51 PM

The domain of the function $f(x) = \sqrt {\frac{{1 + \cos (x)}}{{1 - \cos (x)}}}$ is $\mathbb{R}\backslash \left\{ {2k\pi :k \in \mathbb{Z}} \right\}$ .
See my post above.

**VonNemo19** · June 20th 2009, 02:04 PM

Originally Posted by Plato

The domain of the function $f(x) = \sqrt {\frac{{1 + \cos (x)}}{{1 - \cos (x)}}}$ is $\mathbb{R}\backslash \left\{ {2k\pi :k \in \mathbb{Z}} \right\}$ .
See my post above.

I'm not sure I understand. x is an element of E, not R.

$-\pi\leq{x}\leq\pi\leftarrow{E}$

Perhaps this is over my head... But it seems to me as if you've given a general solution for the function as if it were initially given for all x.

If we reread the problem without all the snazzy language, all it says is find the domain of the function over the interval $[-\pi,\pi]$

**Raoh** · June 20th 2009, 03:05 PM

thanks a lot for everyone

the domain is indeed : $\left [ -\pi,+\pi \right ]\setminus \left \{ 0 \right \}$
i have another question,in some books i found : $-1\leq cos(x)< 1$ which is wrong,am i right ?
thanks

**Plato** · June 20th 2009, 03:33 PM

Originally Posted by Raoh

thanks a lot for everyone

the domain is indeed : $\left [ -\pi,+\pi \right ]\setminus \left \{ 0 \right \}$
i have another question,in some books i found : $-1\leq cos(x)< 1$ which is wrong,am i right ?

Is there a typo in your post?
This always true: $\color{blue}-1\le \cos(x) \le 1$ .

**Raoh** · June 20th 2009, 03:38 PM

their "typo"

thanks.

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