# pre-calc problem involving trig

• Dec 9th 2008, 03:55 PM
DHS1
pre-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 PM
skeeter
$\displaystyle \sin{x} \neq 1$ ... why?

so... $\displaystyle x \neq \, ?$
• Dec 9th 2008, 04:31 PM
DHS1
Quote:

Originally Posted by skeeter
$\displaystyle \sin{x} \neq 1$ ... why?

so... $\displaystyle x \neq \, ?$

I'm sorry but I don't see how you got $\displaystyle \sin{x} \neq 1$

Not saying it's wrong, but can someone tell me how he got that?
• Dec 9th 2008, 04:54 PM
skeeter
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 PM
DHS1
Quote:

Originally Posted by skeeter
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.

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 PM
Professor Fate
Quote:

Originally Posted by DHS1
Am I missing something?

Look again. There's a big difference between the statement $\displaystyle x \neq 1$ and $\displaystyle \sin x \neq 1$
• Dec 9th 2008, 05:18 PM
DHS1
Quote:

Originally Posted by Professor Fate
Look again. There's a big difference between the statement $\displaystyle x \neq 1$ and $\displaystyle \sin x \neq 1$

Wow I can't believe I missed that... haha!

You, sir, have been thanked. (you too skeeter! especially you!)