Trigonometric equations: cotangent

Question

If I have

$cot(x-a)=cot(x-b)$

Where x is in radians and equal on both the sides and not equal to $0$ or $π$ Also for a and b, they are not equal to $0$ or $π$

Does the above equality mean $a=b$? If not then how do we even find the value of a and b? Do we need any more conditions?

There are no singularity points of cotangent between points $(x-a)$ and $(x-b)$

score 4 · Accepted Answer · answered Aug 18 '19 at 09:36

4

Is $a=b$ ?

Not necessarily. Because $\cot$ is periodic with a period of $n\pi , n\in \Bbb Z$.

So, $\cot(x-a) = \cot(x-b) = \cot(n\pi+x-b) \implies \color{purple}{a =m\pi +b} , m =-n \in \Bbb Z$

answered Aug 18 '19 at 09:36

19aksh

sorry I missed to mention something. Can you check the answer for it again? – Korra Aug 18 '19 at 10:05
1

The condition that $\cot(\theta)$ is continuous between $\theta=x-a$ and $\theta=x-b$ rules out all the cases where $m\neq 0,$ so indeed $a=b.$ – David K Aug 18 '19 at 11:18
@David K what if I have $cot(λ(x-b))$ what you said above, does it hold true? Bcz if I have $λ[(x-a)-(x-b)]<π$ it would imply $b<(π/λ)+a$ – Korra Aug 18 '19 at 12:09
If it's $\cot(\lambda(x-a))$ and $\cot(\lambda(x-b))$ then you have $\lambda a = m\pi + \lambda b.$ In that case the continuity condition implies that $\lvert \lambda(b-a)\rvert < \pi,$ so again you have $a=b.$ – David K Aug 18 '19 at 12:18
@David K sorry I don't get it. it is $<π$ why does it imply a=b? We can have $π/3$ in the interval for example. – Korra Aug 18 '19 at 13:09
If $\lambda = 3$ then indeed $a=b+\pi/3$ is a solution to $\cot(\lambda(x-a)) = \cot(\lambda(x-b)).$ But then the condition for continuity is not $\lvert b-a\rvert<\pi,$ it is $\lvert \lambda(b-a)\rvert<\pi.$ If $\lambda = 3$ and $a=b+\pi/3$ then $\lvert \lambda(b-a)\rvert = \lvert 3(b-(b+\pi/3))\rvert = \lvert -\pi\rvert = \pi.$ – David K Aug 18 '19 at 13:17
Let us continue this discussion in chat. – Korra Aug 18 '19 at 13:28

score 3 · Answer 2 · answered Aug 18 '19 at 09:35

3

Hint: Use that $$\cot(x)-\cot(y)=-\csc (x) \csc (y) \sin (x-y)$$

answered Aug 18 '19 at 09:35

?? But I do t want for cot(x)-cot(y)@Dr. Sonnhard Graubner. Thanks for the equation though, might be helpful someway – Korra Aug 18 '19 at 09:39
1

@Korra To make this hint a little more explicit, substitute $x-a$ for $x$ and $x-b$ for $y.$ Then the identity in this answer tells you that $\cot(x-a)-\cot(x-b)=-\csc(x-a)\csc(x-b)\sin(b-a).$ And you already know that $\cot(x-a)-\cot(x-b)=0.$ This tells you something about the right-hand side of the equation, especially $\sin(b-a).$ – David K Aug 18 '19 at 11:16
@David K yup! Got it $sin(b-a)=0 ==> b-a=nπ$ – Korra Aug 18 '19 at 11:26

score 1 · Answer 3 · edited Aug 18 '19 at 10:57

1

$\cot$ is a periodic function that repeats its value after ever $π$ interval i.e $\cot(X) = \cot (nπ + X)$.

edited Aug 18 '19 at 10:57

19aksh

answered Aug 18 '19 at 09:42

3 Answers3