How do you find the exact value of ${tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $tan^-1 (-sqrt3/3)$ ?

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How do you find the exact value of ${tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $tan^-1 (-sqrt3/3)$ ?

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Answer 1 · 2016-10-03T11:32:21

$- \frac{π}{6}$ $-pi/6$

Explanation:

${tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $tan^-1(-sqrt3/3)$

${tan}^{- 1} x$ $tan^-1x$ means find the ANGLE that has a tangent of $x$ $x$

The range of ${tan}^{- 1}$ $tan^-1$ is $- \frac{π}{2}$ $-pi/2$ to $\frac{π}{2}$ $pi/2$

$- \frac{\sqrt{3}}{3}$ $-sqrt3/3$ would fall in the fourth quadrant, so the value of ${tan}^{- 1}$ $tan^-1$ is between $- \frac{π}{2}$ $-pi/2$ and $0$ $0$ and is a negative angle.

Recall the identity $tan x = \frac{sin θ}{cos θ}$ $tanx =sintheta/costheta$

Looking at the unit circle,

$tan (\frac{11 π}{6}) = \frac{sin (\frac{11 π}{6})}{cos (\frac{11 π}{6})} = \frac{- \frac{1}{2}}{\frac{\sqrt{3}}{2}} = - \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$ $tan((11pi)/6)=frac{sin((11pi)/6)}{cos((11pi)/6)}=frac{-1/2}{sqrt3/2}=-1/2*2/sqrt3=-sqrt3/3$

However, because the range of ${tan}^{- 1}$ $tan^-1$ is $\frac{π}{2}$ $pi/2$ to $- \frac{π}{2}$ $-pi/2$ ,
the answer is $- \frac{π}{6}$ $-pi/6$ instead of $\frac{11 π}{6}$ $(11pi)/6$

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How do you find the exact value of ${tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $tan^-1 (-sqrt3/3)$ ?

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How do you find the exact value of ${tan}^{- 1} (- \frac{\sqrt{3}}{3})$ $tan^-1 (-sqrt3/3)$ ?