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How do you find the exact value of #arc tan(-sqrt3)#?

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May 14, 2015

#sin(pi/3) = sqrt(3)/2# and #cos(pi/3) = 1/2#

Since #sin(-theta) = -sin theta# and #cos(-theta) = cos theta# for any #theta#, we have

#tan(-pi/3) = sin(-pi/3)/cos(-pi/3) = (-sin(pi/3))/cos(pi/3)#

#= -(sqrt(3)/2)/(1/2) = -sqrt(3)/1 = -sqrt(3)#

So #arctan(-sqrt(3)) = -pi/3#

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