How do you evaluate #cot[arccos(-1/2)+arccos(0)+arctan(1/sqrt(3))]#?

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Answer 1 · 2016-07-04T15:00:00

#1/sqrt 3#.

A trigonometric function is the same for any general value. So, only

principal values are considered.

#a = arc cos (-1/2). cos a = (-1/2)<0.#

a = angle in the 2nd quadrant.. .

So, a = #(2pi)/3#

#b = arc cos 0 = pi/2#.

#c = arc tan (1/sqrt 3 )= pi/6#.

So, the given expression

#= cot ( a + b + c ) =cot((2pi)/3+ pi/2+pi/6)#

#= cot ((4pi)/3) = cot (pi+pi/3)=cot (pi/3)#

#=1/sqrt 3#.

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