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

by Cuisine123
Tags: calculator, cotx2, scientific, solve
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Cuisine123
#1
Jan20-10, 11:06 PM
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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?
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Dick
#2
Jan20-10, 11:12 PM
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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
#3
Jan21-10, 12:24 AM
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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
#4
Jan21-10, 12:48 AM
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How do I solve for x in cotx=2 by using a scientific calculator?

Quote Quote by Char. Limit View Post
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.
Char. Limit
#5
Jan21-10, 12:56 AM
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Thanks for the proof.
vela
#6
Jan21-10, 01:09 AM
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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
#7
Jan21-10, 01:19 AM
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Aha thanks for that neat little info I didn't know
Bohrok
#8
Jan21-10, 11:32 AM
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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
#9
Jan21-10, 07:14 PM
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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
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Jan21-10, 08:20 PM
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Quote Quote by Mentallic View Post
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.
Char. Limit
#11
Jan21-10, 08:57 PM
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Thanks, Dick. I didn't quite know myself.
Mentallic
#12
Jan22-10, 12:54 AM
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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
#13
Jan22-10, 01:09 AM
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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
#14
Jan22-10, 08:40 AM
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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.


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