Question

Better way to express the domain of sec, csc, and cot functions?

Lol, my textbook writes it in an ugly and confusing way. As an example, they write domain of sec functions like x does not equal pi/2, +/- 3pi/2....

Is there an easier way to express this? Maybe like x does not equal pi/4+(k*pi/2)? Is this correct?

Answer 1

Can someone clarify this for me? I have a test tomorrow lol...

Answer 2

\[secx=\frac{1}{cosx}\]
where is cosx,0?
cosx=0 when x=pi/2, 3pi/2, and so on
whenever we go around a circle that is 2pi
so we can keep going around a circle forever and forever
so cosx=0 when x=pi/2+2npi and x=3pi/2+2npi
for n=...,-3,-2,-1,0,1,2,3,....

Answer 3

Domain of Cot and Cosec - all R except integer multiples of pi
Sec-all R except odd integer multiples of pi/2

Answer 4

Sec is same as Tan

Answer 5

because tan=sin/cos

Answer 6

they have the same denominator

Answer 7

Sorry, I am still confused ><

Answer 8

about?

Answer 9

can you narrow down for me what you are confused about?

Answer 10

The domain of sec x and csc x.. about +2npi

Answer 11

n is the number of times we go around a circle

Answer 12

domain of y=cot x.......x belongs to R-(n pi)
domain of y=cosec x............x belongs to (-infinity,-1]union[1, infinity)
domain of y=sec x.............x belongs to R-(2n+1)pi/2
(here n is a natural number)

Answer 13

we can go counterclockwise or clockwise
that is why we have the negative and the positive n

Answer 14

Oh no, not about the n, about why you would want to add 2npi in general. Like why can't you add it by the period ?

Answer 15

?

Answer 16

Uhm, I mean like so the sec function has asymptotes right? There is an asymptote on -3pi/2, -pi/2, pi/2, 3pi/2. So I understand how my book lists it like that, but I think it is ugly.

Answer 17

\[\cos(\frac{\pi}{2}+2npi)=\cos(\frac{\pi}{2})\cos(2npi)-\sin(\frac{\pi}{2})\sin(2npi)=0(1 or -1)-1(0)=0-0=0\]

Answer 18

\[\cos(\frac{3\pi}{2}+2npi)=\cos(\frac{3\pi}{2})\cos(2npi)-\sin(\frac{3\pi}{2})\sin(2npi)=0(-1 or 1)-1(0)=0-0=0\]

Answer 19

yes we have vertical asymptotes at
\[x=\frac{\pi}{2}+2npi, x=\frac{3\pi}{2}+2npi, n \in \mathbb{Z}\]

Answer 20

What about in 3csc(3x+pi)-2?
So I first factor out the 3 = y = 3csc(3(x+pi/3))-2

I know my period is 2pi/3
So what would be my domain?

Answer 21

\[\csc(3x+\pi)=\frac{1}{\sin(3x+\pi)}\]
we need to figure out when sin(3x+pi) is 0
so remember sin(k) is zero when
\[k=\pi+2npi, k=2\pi+2npi, k \in \mathbb{Z}=> 3x+\pi=2\pi+2npi, 3x+\pi=\pi+2npi\]
now we need to solve for x to find what the domain should exclude

Answer 22

\[x=\frac{\pi+2npi}{3}, x=\frac{2npi}{3}\]

Answer 23

the domain is all real numbers except \[x=\frac{\pi+2npi}{3}, x=\frac{2npi}{3} \]

Answer 24

How did you get this? "x=2npi/3" ? I understand how you got x=pi + 2npi/3..

Answer 25

we have 3x+pi=pi+2npi
subtract pi on both sides
3x=2npi

divide both sides by 3
x=2npi/3