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

If someone says arccot(x) = arctan(1/x) they are assuming the convention that arccot takes values on (0,pi). The other convention is that arccot takes values on (-pi/2,pi/2). In this convention arccot(x) = pi/2-arctan(x) (for x>0). In summary, there are two common conventions for defining arccot and which 'arc' identities are true depends on which convention you are using. In summary, there are two common conventions for defining arccot and the identities for arccot(x) = pi/2-arctan(x) and arccot(x) = arctan(1/x
  • #1
Cuisine123
38
0

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?
 
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  • #2
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
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.
 
  • #4
Char. Limit said:
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.

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.
 
  • #5
Thanks for the proof.
 
  • #6
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
Aha thanks for that neat little info I didn't know :smile:
 
  • #8
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)
 
Last edited:
  • #9
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?
 
  • #10
Mentallic said:
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?

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
Thanks, Dick. I didn't quite know myself.
 
  • #12
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.
 
  • #13
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.
 
  • #14
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.
 

1. How do I enter the equation into the calculator?

To enter the equation into a scientific calculator, you will need to use the "cot" function. This is usually found by pressing the "2nd" or "shift" button, followed by the "tan" button. Then, enter the value of x after the equal sign. For this equation, you would enter "cotx=2" as "cot(x)=2".

2. Where do I find the "inverse" or "x^-1" button on the calculator?

The "inverse" or "x^-1" button is commonly found on scientific calculators as the "2nd" or "shift" function of the "x" button. Press this button to access the inverse function for solving equations.

3. How do I know which mode to use on my calculator?

Most scientific calculators have different modes for calculations, such as "degrees" or "radians". For this equation, you will need to use the "radians" mode. Check your calculator's manual or look for a button labeled "mode" to change the mode.

4. What do I do if the calculator gives me an error?

If the calculator gives you an error, it means that there is no solution for the given equation. This could be due to a mistake in entering the equation or because there is no real solution for the given values.

5. How do I interpret the solution given by the calculator?

The solution given by the calculator will be in radians. If you need the answer in degrees, you can use the conversion function on the calculator. Also, keep in mind that there may be multiple solutions for the given equation, so the calculator may only give you one possible solution.

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