arctan = 1/tan right?

minendo

Elite Member
Aug 31, 2001
35,560
22
81
yeah

actually i dont know cause its been years since i have taken trig, but i do remember the book being helpful
 

josphII

Banned
Nov 24, 2001
1,490
0
0


<< Technically, it's the integral of 1/(1+x^2) with respect to x. >>



actually its an infinite series, as is pi, e, sin, cos, etc
 

DanFungus

Diamond Member
Jul 27, 2001
5,857
0
0
arctan uses the same range as Sin, so it can be from 90 degrees (pi/2) to -90 degrees (-pi/2)
 

StandardCell

Senior member
Sep 2, 2001
312
0
0


<<

<< Technically, it's the integral of 1/(1+x^2) with respect to x. >>



actually its an infinite series, as is pi, e, sin, cos, etc
>>



Uhhh, it's BOTH...the Taylor series expansion is a way of expressing it explicitly. I have merely expressed it compactly.
 

Ionizer86

Diamond Member
Jun 20, 2001
5,292
0
76
the notation tan^-1 (X) is simply arctan (X). This is odd, since tan^2 (X) is actually (tan X)^2

I dislike the notation, but anyway, arctan X (or tan^-1 (X)) helps you to go from angle measure back to side proportion. For example, tan (30) (or tan pi/6 for you radian lovers) is equal to 1/2, since the opposite side will be half the length of the adjacent side.

Now, arctan (1/2) is 30 (or pi/6) because this brings you from opposite/adjacent side proportion back to the angle measure.
 

HappyFace

Diamond Member
Nov 2, 1999
6,265
5
81
:confused: Please stop all this math stuff before I fall over dead and puke my intestines all over the carpet.