# Thread: asymptotes of a cot function.

1. ## asymptotes of a cot function.

I had a question similar to this yesterday but I cant figure this one out.
what are the equations of the asymptotes for the function
y=(3)(cot)(3)(pi)(x)

y=
3 sin 3 pi x
3 cos 3pi x

one of the asympotes is found by making the bottom = 0

so I get

x cannot be apx. = .1667

but how can I find the equation for the asymptotes from that?
answer is either

x=n
x=n/2
x=n/3
x=2n
x=3n
thanks for any help.

2. Originally Posted by brentwoodbc
I had a question similar to this yesterday but I cant figure this one out.
what are the equations of the asymptotes for the function
y=(3)(cot)(3)(pi)(x)

y=
3 sin 3 pi x
3 cos 3pi x

one of the asympotes is found by making the bottom = 0

so I get

x cannot be apx. = .1667

but how can I find the equation for the asymptotes from that?
answer is either

x=n
x=n/2
x=n/3
x=2n
x=3n
thanks for any help.
cot A = cos A/sin A. NOT sin A/cos A. So you've started with the wrong expression.

Solve 3 sin (3pi x) = 0 => sin (3pi x) = 0

=> 3pi x = n pi where n is any integer.

Solve for x.

3. Originally Posted by The Second Solution
cot A = cos A/sin A. NOT sin A/cos A. So you've started with the wrong expression.

Solve 3 sin (3pi x) = 0 => sin (3pi x) = 0

=> 3pi x = n pi where n is any integer.

Solve for x.
Im a little confused.
how do you get 3pi x = n pi?
Just from the period?

and how come you divided out the 3sin? I mean cant you divide out the whole thing other than the x?

anyways I get x = n/3,

4. Originally Posted by brentwoodbc
Im a little confused.
how do you get 3pi x = n pi?
Just from the period?

and how come you divided out the 3sin? I mean cant you divide out the whole thing other than the x?

anyways I get x = n/3,
Both sides of the equation were divided by 3.

The basic fact that if sin(A) = 0 then A = n pi has been used.