# Homework Help: Cot^(-1)(0) = pi/2, why?

1. Aug 4, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

I don't understand why cot^(-1)(0) = pi/2 and was hoping someone could explain this to me. cot(theta)=1/tan(theta)
because tan^(-1)(0) is undefined
2. Relevant equations

3. The attempt at a solution

2. Aug 4, 2011

### Pengwuino

I think you're confusing $cot^{-1}(\theta) = {{1}\over{cot(\theta)}}$ with the Arc-cotangent or "Inverse cotangent", which is the inverse function of $cotan(\theta)$. The inverse cotangent is the function that tells you what values of $\theta$ give you the value 'x' in $cotan(\theta) = x$.

So in other words, $cot^{-1}(0)$ is asking what values of $\theta$ give you $cot(\theta) = 0$.

EDIT: Ok I think I got that squared away correctly.

Last edited: Aug 4, 2011
3. Aug 4, 2011

### GreenPrint

oh so cot(0) is undefined because
cot(0) = 1/tan(0) = 1/0 = undefined, makes sense
cot(pi/2) = 1/tan(pi/2) = 1/undefined =/= 0
I don't see how the two are equal, and ya I think I may be getting some things mixed up as I haven't dealt with basic trig in several years lolz

4. Aug 4, 2011

### Pengwuino

No, cot(pi/2) = 0.

Why would they be equal in the first place?

5. Aug 4, 2011

### GreenPrint

cot(theta)=1/tan(theta)
cot(pi/2) should then be equal to 1/tan(pi/2)
cot(pi/2) = 0 = 1/tan(pi/2)

I just don't understand why
1/tan(pi/2) is equal to zero
because tan(pi/2) = undefined
so 1/tan(pi/2) = 1/undefined
how is this equal to zero?[/quote]
Rather than think of cot(x) as 1/tan(x), a more fundamental definition is: tan(x)= sin(x)/cos(x) and cot(x)= cos(x)/sin(x). $sin(\pi/2)= 1$ and $cos(\pi/2)= 0$ so $tan(\pi/2)$ is undefined (the denominator is 0) while $cot(\pi/2)= 0$ (the numerator is 0).

Last edited by a moderator: Aug 5, 2011
6. Aug 4, 2011

### SammyS

Staff Emeritus
cot-1(x) is another way to write the arccot(x) function.

cot-1(x) ≠ 1/cot(x) .

7. Aug 5, 2011

### Mentallic

It's undefined in the sense that division by zero is not allowed. Think about it this way, if $$\frac{a}{\left(\frac{b}{c}\right)}=\frac{ac}{b}$$ then $$\frac{1}{\left(\frac{1}{0}\right)}=0$$

Or you can even think of division by zero as being $\pm\infty$ so when we divide a finite value by this amount, we get 0.

8. Aug 5, 2011

### uart

Hi Greenprint. Mentallic has the correct answer here. $1/0$ is undefined however $0/1$ is perfectly well defined and is equal to zero.

9. Aug 6, 2011

### Bohrok

$\cot^{-1}(0) = \pi/2 \Rightarrow \cot(\pi/2) = 0$

Trying to rearrange it so you can use the more familiar tan doesn't really help since it brings in division by 0. Take a look at the graph of cot-1x and see what the value is when x = 0

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10. Aug 6, 2011

### HallsofIvy

I think it would help to point out that the basic definition of $cot(x)$ is $cos(x)/sin(x)$, not $1/tan(x)$. At $x= \pi/2$, $tan(x)$ is not defined but $cos(\pi/2)= cos(\pi/2)/sin(\pi/2)= 0/1= 0$

11. Aug 6, 2011

### SammyS

Staff Emeritus
O.P. acknowledged that $\cot(\pi/2)= 0$ in Post #5.