How do you find the exact value of #tan^-1 (-sqrt3/3)#?

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Answer 1 · 2016-10-03T11:32:21

#-pi/6#

#tan^-1(-sqrt3/3)#

#tan^-1x# means find the ANGLE that has a tangent of #x#

The range of #tan^-1# is #-pi/2# to #pi/2#

#-sqrt3/3# would fall in the fourth quadrant, so the value of #tan^-1# is between #-pi/2# and #0# and is a negative angle.

Recall the identity #tanx =sintheta/costheta#

Looking at the unit circle,

#tan((11pi)/6)=frac{sin((11pi)/6)}{cos((11pi)/6)}=frac{-1/2}{sqrt3/2}=-1/2*2/sqrt3=-sqrt3/3#

However, because the range of #tan^-1# is #pi/2# to #-pi/2#,
the answer is #-pi/6# instead of #(11pi)/6#