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How do I solve for x in cotx=2 by using a scientific calculator? 
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#1
Jan2010, 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? 


#2
Jan2010, 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.



#3
Jan2110, 12:24 AM

PF Gold
P: 1,941

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
Jan2110, 12:48 AM

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


#6
Jan2110, 01:09 AM

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PF Gold
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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 cofunctions.



#7
Jan2110, 01:19 AM

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



#8
Jan2110, 11:32 AM

P: 867

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) 


#9
Jan2110, 07:14 PM

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Well the trig identity was derived from the properties of a righttriangle 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 nonzero x? 


#10
Jan2110, 08:20 PM

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#12
Jan2210, 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/2tan^{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
Jan2210, 01:09 AM

PF Gold
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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. 


#14
Jan2210, 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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