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

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?

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.

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...

[tex]arccot(x)=\frac{\pi}{2}-arctan(x)[/tex]

[tex]arccot(x)=arctan(\frac{1}{x}), x>0[/tex]

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

Mentallic

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

You can prove the second result by a similar method.

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

Char. Limit

Thanks for the proof.

vela

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

Mentallic

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

Bohrok

Other proofs that arccot(x) = [itex]\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[/itex]/2 - θ) (a trig identity)
x = cot([itex]\pi[/itex]/2 - θ)
arccot(x) = [itex]\pi[/itex]/2 - θ
tan(θ) = x
θ = arctan(x)
arccot(x) = [itex]\pi[/itex]/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)

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 [tex]arccot(x)=arctan(\frac{1}{x}), x>0[/tex] as char.limit has posted. Why isn't this true for all non-zero x?

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.

Char. Limit

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

Mentallic

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

[tex]cot^{-1}x=\pi/2-tan^{-1}x[/tex]
[tex]cot^{-1}x=tan^{-1}(1/x)[/tex]

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.

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.

Arctangent I don't know about.

Dick

I've just realized there is a another common convention for defining arccot. You can also define it to take values on (-pi/2,pi/2) but you pay the price of having it be discontinuous at 0. http://mathworld.wolfram.com/InverseCotangent.html Which 'arc' identities are true depends on what convention you are using. You HAVE to say which.