How do you evaluate ${tan}^{- 1} (- (\frac{\sqrt{3}}{3}))$ $tan^-1(-(sqrt(3)/3))$ ?

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Answer 1 · 2016-02-20T05:13:39

We notice that for an angle $θ$ $theta$ we have that

$tan θ = - \frac{\sqrt{3}}{3} = - \frac{1}{\sqrt{3}} = - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$ $tantheta=-sqrt3/3=-1/(sqrt3)=-(1/2)/(sqrt3/2)$

Now we need to find the angles $θ_{1}$ $theta_1$ , $θ_{2}$ $theta_2$ which have

${sin θ}_{1} = - \frac{1}{2}$ $sintheta_1=-1/2$ and ${cos θ}_{1} = \frac{\sqrt{3}}{2}$ $costheta_1=sqrt3/2$

and

${sin θ}_{2} = \frac{1}{2}$ $sintheta_2=1/2$ and ${cos θ}_{2} = - \frac{\sqrt{3}}{2}$ $costheta_2=-sqrt3/2$

These angles are $θ_{1} = 150^{o} = 5 \cdot \frac{π}{6}$ $theta_1=150^o=5*pi/6$ and $θ_{2} = 330^{o} = 11 \cdot \frac{π}{6}$ $theta_2=330^o=11*pi/6$

How do you evaluate tan−1(−(√33))tan^-1(-(sqrt(3)/3))?