When is cot(x) defined?Watch

#1
I know we take cot(x) as undefined if and only if (when x is real) sin(x)=0, since we have cot(x)=cos(x)/sin(x), and so we have asymptotes only at the values of x for which sin(x)=0.
But we define cot(x)=1/tan(x), so surely tan(x) must be defined for cot(x) to be defined? By that reasoning, since tan(x)=sin(x)/cos(x), cot(x) is undefined when cos(x)=0, but this is apparently incorrect as there are no asymptotes on the graph at those x values for which cos(x)=0?
This has been confusing me for a long time so if anyone can clear things up that would be greatly appreciated .
0
4 years ago
#2
(Original post by PrimeLime)
I know we take cot(x) as undefined if and only if (when x is real) sin(x)=0, since we have cot(x)=cos(x)/sin(x), and so we have asymptotes only at the values of x for which sin(x)=0.
But we define cot(x)=1/tan(x), so surely tan(x) must be defined for cot(x) to be defined? By that reasoning, since tan(x)=sin(x)/cos(x), cot(x) is undefined when cos(x)=0, but this is apparently incorrect as there are no asymptotes on the graph at those x values for which cos(x)=0?
This has been confusing me for a long time so if anyone can clear things up that would be greatly appreciated .
cotx has not got a finite value at Nπ, where N is an integer
0
4 years ago
#3
(Original post by PrimeLime)
I know we take cot(x) as undefined if and only if (when x is real) sin(x)=0, since we have cot(x)=cos(x)/sin(x), and so we have asymptotes only at the values of x for which sin(x)=0.
But we define cot(x)=1/tan(x), so surely tan(x) must be defined for cot(x) to be defined? By that reasoning, since tan(x)=sin(x)/cos(x), cot(x) is undefined when cos(x)=0, but this is apparently incorrect as there are no asymptotes on the graph at those x values for which cos(x)=0?
This has been confusing me for a long time so if anyone can clear things up that would be greatly appreciated .
If we write cot(x) = 1/tan(x), we clearly see that cot(x) is not defined if and only if tan(x)=0 (otherwise we'd be dividing by 0)

When tan(x) = 0, x = npi (for any integer n)
Thus, these are the only values for which cot(x) is not defined.

I hope that helps.
0
4 years ago
#4
(Original post by PrimeLime)
I know we take cot(x) as undefined if and only if (when x is real) sin(x)=0, since we have cot(x)=cos(x)/sin(x), and so we have asymptotes only at the values of x for which sin(x)=0.
But we define cot(x)=1/tan(x),
so cot(x) is undefined if and only if tan(x)=0.
Since tan(x)=sin(x)/cos(x), tan(x)=0 if and only if sinx=0 as you wrote above.

For cos(x)=0, tan(x) is undefined (infinity) so cot(x)=0 (defined) there
0
#5
(Original post by razzor)
If we write cot(x) = 1/tan(x), we clearly see that cot(x) is not defined if and only if tan(x)=0 (otherwise we'd be dividing by 0)

When tan(x) = 0, x = npi (for any integer n)
Thus, these are the only values for which cot(x) is not defined.

I hope that helps.
But cot(x)=1/tan(x) means that tan(x)=sin(x)/cos(x) must be defined, i.e. cos(x) cannot be equal to 0 either.
Basically, neither sin(x) nor cos(x) can be equal to 0, so we deduce that:
cot(x) is undefined for x=(kpi)/2.
The problem is that the graph has asymptotes only at INTEGER multiples of pi (and not the half-integer multiples), which contradicts our deduction???
0
#6
(Original post by ztibor)
so cot(x) is undefined if and only if tan(x)=0.
Since tan(x)=sin(x)/cos(x), tan(x)=0 if and only if sinx=0 as you wrote above.

For cos(x)=0, tan(x) is undefined (infinity) so cot(x)=0 (defined) there
The only other people I've told this problem to have presented the same argument as you have: cos(x)=0 implies that tan(x) can be taken as infinity so that 1/tan(x)=cot(x)=0. But this is nonsense since if tan(x) is undefined then 1/tan(x) cannot evaluated and is undefined. There's no, "if tan(x) is undefined we can call it infinity so that 1/infinity=0". That's not maths.

Sorry if that sounded a bit blunt by the way. I didn't mean to insult your maths; I'm just saying that no one seems to be able to provide a valid answer.
0
4 years ago
#7
(Original post by PrimeLime)
But cot(x)=1/tan(x) means that tan(x)=sin(x)/cos(x) must be defined, i.e. cos(x) cannot be equal to 0 either.
Basically, neither sin(x) nor cos(x) can be equal to 0, so we deduce that:
cot(x) is undefined for x=(kpi)/2.
The problem is that the graph has asymptotes only at INTEGER multiples of pi (and not the half-integer multiples), which contradicts our deduction???
Let us choose an x value for which tan(x) is undefined (say x=pi/2).

Then at this point, tan(x) is equal to infinity (just look at the graph of tan).

At this point, cot(x) becomes 1/infinity.
limx->infinity 1/x is 0, so we can say that when tan(x) is undefined, cot(x) = 0. Thus, cot(x) is indeed defined at that point.
0
4 years ago
#8

For me, cot(x) = cos(x)/sin(x)

so it's only undefined when sin(x) = 0

You could use the similar definition cot(x) = 1/tan(x) but that doesn't quite work for the reason you state. So if you wanted to work this way, you would need to extend the definition of cot(x) so that it has the value cot(x) = 0 when tan(x) is undefined. That can be called 'extending' the function. It makes cot continuous at those points so it makes sense.

Does this help?
1
#9
(Original post by razzor)
Let us choose an x value for which tan(x) is undefined (say x=pi/2).

Then at this point, tan(x) is equal to infinity (just look at the graph of tan).

At this point, cot(x) becomes 1/infinity.
limx->infinity 1/x is 0, so we can say that when tan(x) is undefined, cot(x) = 0. Thus, cot(x) is indeed defined at that point.
tan(x) is not "equal to infinity" though. That's impossible. Yes, towards pi/2 it TENDS towards infinity, but you must take it as value-less at pi/2. So that would make cot(x) value-less.
0
#10
(Original post by ian.slater)

For me, cot(x) = cos(x)/sin(x)

so it's only undefined when sin(x) = 0

You could use the similar definition cot(x) = 1/tan(x) but that doesn't quite work for the reason you state. So if you wanted to work this way, you would need to extend the definition of cot(x) so that it has the value cot(x) = 0 when tan(x) is undefined. That can be called 'extending' the function. It makes cot continuous at those points so it makes sense.

Does this help?
This makes a lot of sense.
The only problem that exists now is everyone defining cot(x)=1/tan(x) and leaving it there. I tried telling other people that cot(x) does not equal 1/tan(x) but it didn't go down well XD.
Thanks for your help and to everyone else .

EDIT: To sum up the problem, I have realised that a/b is not necessarily equal to 1(b/a). If a=0 then a/b=0 but 1/(b/a) is undefined.
0
4 years ago
#11
(Original post by PrimeLime)
I know we take cot(x) as undefined if and only if (when x is real) sin(x)=0, since we have cot(x)=cos(x)/sin(x), and so we have asymptotes only at the values of x for which sin(x)=0.
But we define cot(x)=1/tan(x), so surely tan(x) must be defined for cot(x) to be defined? By that reasoning, since tan(x)=sin(x)/cos(x), cot(x) is undefined when cos(x)=0, but this is apparently incorrect as there are no asymptotes on the graph at those x values for which cos(x)=0?
This has been confusing me for a long time so if anyone can clear things up that would be greatly appreciated .
Its still continuous if tanx->infinty and takes a finit value. Don't understand ur problem

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0
4 years ago
#12
(Original post by physicsmaths)
Its still continuous if tanx->infinty and takes a finit value. Don't understand ur problem

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OP has correctly raised a good point. At school level you can wave a few hands and graphs and claim that since tan(x) gets very big as x -> pi/2 then 1/tan(x) -> 0 as x -> pi/2. So cot(pi/2) = 1/tan(pi/2) = 0.

OP is more rigourous

He says "although the limit exists, tan(pi/2) isn't defined. So 1/tan(pi/2) isn't defined. And so this definition can't hold for cot(pi/2)". OP is applying a first-year Uni level of rigour. Which is great for STEP thought, but can get you into trouble if teachers say "I don't see your problem".
0
4 years ago
#13
(Original post by ian.slater)
OP has correctly raised a good point. At school level you can wave a few hands and graphs and claim that since tan(x) gets very big as x -> pi/2 then 1/tan(x) -> 0 as x -> pi/2. So cot(pi/2) = 1/tan(pi/2) = 0.

OP is more rigourous

He says "although the limit exists, tan(pi/2) isn't defined. So 1/tan(pi/2) isn't defined. And so this definition can't hold for cot(pi/2)". OP is applying a first-year Uni level of rigour. Which is great for STEP thought, but can get you into trouble if teachers say "I don't see your problem".
Kool

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0
#14
(Original post by ian.slater)
OP has correctly raised a good point. At school level you can wave a few hands and graphs and claim that since tan(x) gets very big as x -> pi/2 then 1/tan(x) -> 0 as x -> pi/2. So cot(pi/2) = 1/tan(pi/2) = 0.

OP is more rigourous

He says "although the limit exists, tan(pi/2) isn't defined. So 1/tan(pi/2) isn't defined. And so this definition can't hold for cot(pi/2)". OP is applying a first-year Uni level of rigour. Which is great for STEP thought, but can get you into trouble if teachers say "I don't see your problem".
Haha, I've only just finished AS Maths but yes, I do try to be as rigorous as possible since I have to do STEP and am trying to get to Cambridge.
I have actually found a lot of other problems like this, but most other people don't understand them...
Just out of curiosity, at what level do you study maths currently (if at all)?
0
4 years ago
#15
(Original post by PrimeLime)
Just out of curiosity, at what level do you study maths currently (if at all)?
Check out my profile; I'm a maths tutor who works mostly with FM students or STEP.
0
4 years ago
#16
Either is a perfectly legit definition to choose. Probably best to just state what you're using as your definition (including your domain of definition) and go from there.
0
#17
(Original post by ian.slater)
Check out my profile; I'm a maths tutor who works mostly with FM students or STEP.
Churchill college! That's exactly where I'm applying for maths! XD
0
4 years ago
#18
(Original post by PrimeLime)
Churchill college! That's exactly where I'm applying for maths! XD
Good luck! I've just used your question (slightly modified) in a practice interview for an Oxford hopeful. Sketch cot(x) - give me a definition - give me another - which is better and why? She cracked it after one small hint. Thanks!
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