I need to determine the domain of each function.

If you answer can you tell me how you did it?

$\displaystyle h(x) = \frac{1}{1-sin(x)}$

Thanks.

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- Dec 9th 2008, 03:55 PMDHS1pre-calc problem involving trig
I need to determine the domain of each function.

If you answer can you tell me how you did it?

$\displaystyle h(x) = \frac{1}{1-sin(x)}$

Thanks. - Dec 9th 2008, 04:27 PMskeeter
$\displaystyle \sin{x} \neq 1$ ... why?

so... $\displaystyle x \neq \, ?$ - Dec 9th 2008, 04:31 PMDHS1
- Dec 9th 2008, 04:54 PMskeeter
look at the denominator ...

are you familiar with the rule that says division by 0 is undefined?

if $\displaystyle sin{x} = 1$, what would be the value of the denominator?

so ... what does that tell you $\displaystyle \sin{x}$ cannot be equal to ?

domain is the set of all x-values that can be input into a function without breaking certain rules ... division by 0 is one of those rules that cannot be broken. - Dec 9th 2008, 05:05 PMDHS1
Yes I am familiar with that rule. I can see the answer being $\displaystyle x \neq 1$ in an equation that looks like this:

$\displaystyle h(x) = \frac{1}{1-x}$

but if I plug 1 in for x on my problem below, the denominator looks like this:

$\displaystyle h(x) = \frac{1}{1-sin(1)}$ = $\displaystyle \frac{1}{1-0.84147...}$

because sin(1) = 0.84147...

Am I missing something? - Dec 9th 2008, 05:09 PMProfessor Fate
- Dec 9th 2008, 05:18 PMDHS1