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

by Cuisine123
Tags: calculator, cotx2, scientific, solve
 P: 39 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?
 HW Helper Sci Advisor Thanks P: 24,421 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.
 PF Patron P: 1,930 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.
HW Helper
P: 3,320

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

 Quote by 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.
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.
 PF Patron P: 1,930 Thanks for the proof.
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,855 In fact, the "co" in "cosine" is there because the function gives the sine of the COmplementary angle. Same with the other co-functions.
 HW Helper P: 3,320 Aha thanks for that neat little info I didn't know
 P: 867 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)
 HW Helper P: 3,320 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?
HW Helper
 Quote by 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?
 HW Helper P: 3,320 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.