# How do I solve for x in cotx=2 by using a scientific calculator?

1. Jan 20, 2010

### Cuisine123

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

How do I solve for x in cotx=2 by using a scientific calculator?
What do I have to enter?

2. Relevant equations

N/A

3. The attempt at a solution

I know that I can enter tan^-1 (1/2) to find x, but is there another way to do it on a calculator?

2. Jan 20, 2010

### Dick

Why do you need another way? If cot(x)=2 then 1/2=1/cot(x)=tan(x). The only other way to do it is to lobby the calculator companies for a cot^(-1) button.

3. Jan 21, 2010

### Char. Limit

There is another way:

The arccotangent of x is equal to one half pi minus the arctangent of x. The arccotangent of x only equals the arctangent of 1/x if x is greater than zero.

In short...

$$arccot(x)=\frac{\pi}{2}-arctan(x)$$

$$arccot(x)=arctan(\frac{1}{x}), x>0$$

I don't know why this is true, my calculus book tells me so.

4. Jan 21, 2010

### Mentallic

Draw a right triangle with an angle $\theta$ and let the opposite side length be x and the adjacent side be 1 such that $tan\theta=x$. Now, $\theta=arctan(x)$ and if we find the other angle in the triangle in terms of $\theta$, by sum of angles in a triangle, it is $\pi/2 -\theta$, so $tan(\pi/2-\theta)=1/x$ thus $arccot(x)=\pi/2-\theta=\pi/2-arctan(x)$.

You can prove the second result by a similar method.

@ the OP: you don't need an $cot^{-1}$ function on your calculator since it's very simple to take the $tan^{-1}$ of the reciprocal.

5. Jan 21, 2010

### Char. Limit

Thanks for the proof.

6. Jan 21, 2010

### vela

Staff Emeritus
In fact, the "co" in "cosine" is there because the function gives the sine of the COmplementary angle. Same with the other co-functions.

7. Jan 21, 2010

### Mentallic

Aha thanks for that neat little info I didn't know

8. Jan 21, 2010

### Bohrok

Other proofs that arccot(x) = $\pi[/tex]/2 - arctan(x) and arccot(x) = arctan(1/x): arccot(x) = [itex]\pi[/tex]/2 - arctan(x) Let x = tan(θ) = cot([itex]\pi$/2 - θ) (a trig identity)
x = cot($\pi$/2 - θ)
arccot(x) = $\pi$/2 - θ
tan(θ) = x
θ = arctan(x)
arccot(x) = $\pi$/2 - arctan(x)

arccot(x) = arctan(1/x)

Let θ = arccot(x)
cot(θ) = x
1/cot(θ) = tan(θ) = 1/x
θ = arctan(1/x)
arccot(x) = arctan(1/x)

Last edited: Jan 21, 2010
9. Jan 21, 2010

### Mentallic

Well the trig identity was derived from the properties of a right-triangle so your proof hasn't really changed much, other than giving someone the opportunity to realize it can be done that way too.

I'm curious as to why $$arccot(x)=arctan(\frac{1}{x}), x>0$$ as char.limit has posted. Why isn't this true for all non-zero x?

10. Jan 21, 2010

### Dick

arctan and arccot are defined with different angular ranges. Look it up and draw a graph. They only overlap for x>0 or resulting angle between 0 and pi/2.

11. Jan 21, 2010

### Char. Limit

Thanks, Dick. I didn't quite know myself.

12. Jan 22, 2010

### Mentallic

Sorry I didn't quit understand what "angular ranges" meant.
I've drawn a graph for both:

$$cot^{-1}x=\pi/2-tan^{-1}x$$
$$cot^{-1}x=tan^{-1}(1/x)$$

and it seems as though the first is only true for x>0 while the second is true for all x.

Please elaborate so we can settle this.

13. Jan 22, 2010

### Char. Limit

Well, I believe that invtrig functions are only defined for certain angles. For example, I'm pretty sure that arcsin(x) is only defined from -pi/2 to pi/2 on the y axis. Thus, the angular range of arcsin(x) is [-pi/2,pi/2], but don't quote me on that.