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 $(11pi)/6$

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