Why does cotangent inverse of 0 equal pi/2?

[GreenPrint]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[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.

 

[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 you I think I may be getting some things mixed up as I haven't dealt with basic trig in several years lolz

 

[Pengwuino]

No, cot(pi/2) = 0.

Why would they be equal in the first place?

 

[GreenPrint]

No, cot(pi/2) = 0.

Why would they be equal in the first place?

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).

 

[SammyS]

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

cot^-1(x) ≠ 1/cot(x) .

 

[Mentallic]

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?

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.

 

[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.

 

[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

 

[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

 

[SammyS]

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