In radians, what is #arccos-1/2#?

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Answer 1 · 2018-07-27T14:44:38

#{2\pi}/3#

Notice: #0\le \cos^{-1}x\le \pi#

#\therefore \cos^{-1}(-1/2)#

#=\pi-\cos^{-1}(1/2)#

#=\pi-\pi/3#

#={2\pi}/3#

Answer 2 · 2018-07-27T14:57:23

As below.

#theta = arccos (-1/2)#

#cos theta = -1/2 = cos (120) =cos ( (2pi) / 3)^c # or # ((4pi)/3)^c#

Generalizing, #theta = (n pi - ((pi)/3) , (n pi + (pi/3))# where n is an odd integer.