# Domain of trig functions

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1. Mar 30, 2016

### Astraithious

1. The problem statement, all variables and given/known data
Find the domain of this function and check with your graphing calculator:
f(x)=(1+cosx)/(1-cos2x)
2. Relevant equations

3. The attempt at a solution
i get to (1+cosx)/(1+cosx)(1-cosx) which is factored. so then setting each one to zero one at a time i figure out that
cosx = -1 and cosx = 1
then i get stuck from there.

here is what the page says to do

Step 4
Setting each of those factors to zero, we get:

and
Step 5
Solving each of those, we get:

and , where n is an integer
Step 6
Combining those, we get:

, where n is an integer
Step 7
Since those values are where the denominator is zero, our domain is at every value EXCEPT those:

, where n is an integer

I was hoping somebody could break down the steps after 4 , i have no clue where
OR come from, thank you

2. Mar 30, 2016

### SteamKing

Staff Emeritus
You mean you don't know for which angles the cosine function is ±1 ?

Haven't you ever seen a graph of cosine (x) plotted out:

3. Mar 30, 2016

### Ray Vickson

Really? You don't know what are the $\cos$ values of $0$ or or $\pi = 180^{o}$ or of $2 \pi = 360^{o}$? You don't need a calculator to tell you those values; just draw a unit circle and use the geometric definition of $\cos(\theta)$.

4. Mar 30, 2016

### Math_QED

Cos(a) = 1
<=> Cos(a) = cos(0)
<=> a = n2π
n ∈ Z (integer numbers)

Same for cos(a) = -1

In general: cos(x) = cos(z) <=> x = z +n2π or x= -z + n2π

5. Mar 30, 2016

### HallsofIvy

I would NOT do that factoring and canceling because $\frac{1-x}{(1-x)(1+ x)}$ and $\frac{1}{1+ x)}$ do NOT have the same domain. The first, $\frac{1-x}{(1-x)(1+ x}$, has domain "all real numbers except x= 1 and x= -1" while the second, $\frac{1}{1+ x}$, has domain "all real numbers except x= -1".

Instead, I would use the fact that $1- cos^2(x)= sin^2(x)$ so that $\frac{1+ cos(x)}{1- cos^2(x)}= \frac{1+ cos(x)}{sin^2(x)}$. That is defined for all x except those such that sin(x)= 0.

6. Mar 31, 2016

### micromass

Nobody is cancelling anything in this thread.

7. Apr 18, 2016

### UnivMathProdigy

Using this method, wouldn't that leave us with $\frac{2}{0}$ or the indeterminate form $\frac{0}{0}$ given $x=n\pi \quad \forall \quad n \in \mathbb{Z}$?

I also echo Ray Vickson's concern about remembering those trig values. It took me a while to get it, but it shouldn't be that hard to remember.