Homework Help: 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.