# Thread: Help me find Arccot please!

1. ## Help me find Arccot please!

I am trying to solve a question involving inding the Arccot. I've inserted a link and an image file. It's my first time doing this...hope it works. http://www.mathhelpforum.com/math-he...ntid=108&stc=1.

2. Is that correct, sqrt(3/3) ?
Why, sqrt(3/3) is sqrt(1). And sqrt(1) is +,-1.
So, arccot[sqrt(3/3)] = arccot(+,-1)
And so, arccot[-sqrt(3/3)] = arccot[-(+,-1)] = arccot(-,+1) = arccot(+,-1)

Cotangent (and Tangent) function is positive in the 1st and 3rd quadrants,
So, Arccot (1) is 45 or 225 degrees. Or, pi/4 and 5pi/4 radians.

Cotangent (and Tangent) function is negative in the 2nd and 4th quadrants,
So, Arccot (-1) is 135 or 315 degrees. Or, 3pi/4 and 7pi/4 radians.

Therefore, arccot[-sqrt(3/3)] = pi/4, 3pi/4, 5pi/4, or 7pi/4 radians in the interval [0,2pi]. -------answer.

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However, I doudt if that is really sqrt(3/3)....

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I think that is arccot[- (1/3)sqrt(3)]

(1/3)sqrt(3) is actully 1/sqrt(3) if you rationalize the denominator.

Hence the reference angle is 60 degrees or pi/3 radians,
because cot(60deg) = 1/sqrt(3)

Then, arccot[-(1/3)sqrt(3)] -----in the 2nd or 4th quadrants,
= arccot[-1/sqrt(3)]
= 120 or 300 degrees