How do you find the exact value of $a r c tan (- \sqrt{3})$ $arc tan(-sqrt3)$ ?

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How do you find the exact value of $a r c tan (- \sqrt{3})$ $arc tan(-sqrt3)$ ?

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Answer 1 · 2015-05-14T21:10:13

$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3) = sqrt(3)/2$ and $cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3) = 1/2$

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

$tan (- \frac{π}{3}) = \frac{sin (- \frac{π}{3})}{cos (- \frac{π}{3})} = \frac{- sin (\frac{π}{3})}{cos (\frac{π}{3})}$ $tan(-pi/3) = sin(-pi/3)/cos(-pi/3) = (-sin(pi/3))/cos(pi/3)$

$= - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = - \frac{\sqrt{3}}{1} = - \sqrt{3}$ $= -(sqrt(3)/2)/(1/2) = -sqrt(3)/1 = -sqrt(3)$

So $arctan (- \sqrt{3}) = - \frac{π}{3}$ $arctan(-sqrt(3)) = -pi/3$

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How do you find the exact value of $a r c tan (- \sqrt{3})$ $arc tan(-sqrt3)$ ?

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