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

[Cuisine123]

Homework Statement

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

Homework Equations

N/A

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...

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

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

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

 

[Mentallic]

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.

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.

 

[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):<br /> <br /> arccot(x) = [itex]\pi[/tex]/2 - arctan(x)<br /> <br /> Let x = tan(θ) = cot($\pi$/2 - θ) (a trig identity)<br /> x = cot($\pi$/2 - θ)<br /> arccot(x) = $\pi$/2 - θ<br /> tan(θ) = x<br /> θ = arctan(x)<br /> arccot(x) = $\pi$/2 - arctan(x)<br /> <br /> <br /> arccot(x) = arctan(1/x)<br /> <br /> Let θ = arccot(x)<br /> cot(θ) = x<br /> 1/cot(θ) = tan(θ) = 1/x<br /> θ = arctan(1/x)<br /> arccot(x) = arctan(1/x)[/itex][/itex]

 

[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?

 

[Dick]

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?

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:

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

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.

 

[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.