How do you convert $(- \sqrt{3}, - \sqrt{3})$ $(-sqrt(3),-sqrt(3))$ to polar form?

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Answer 1 · 2016-05-04T16:21:11

$(- \sqrt{3}, - \sqrt{3})$ $(-sqrt3,-sqrt3)$ in polar form is $(\sqrt{6}, \frac{π}{4})$ $(sqrt6,pi/4)$

If $(r, θ)$ $(r,theta)$ is in polar form and $(x, y)$ $(x,y)$ in Cartesian form the relation between them is as follows:

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ , $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $tan θ = \frac{y}{x}$ $tantheta=y/x$

As in Cartesian form, the point is $(- \sqrt{3}, - \sqrt{3})$ $(-sqrt3,-sqrt3)$ ,

in polar form, $r = \sqrt{{(- \sqrt{3})}^{2} + {(- \sqrt{3})}^{2}} = \sqrt{3 + 3} = \sqrt{6}$ $r=sqrt((-sqrt3)^2+(-sqrt3)^2)=sqrt(3+3)=sqrt6$

and $tan θ - \frac{- \sqrt{3}}{- \sqrt{3}} = 1$ $tantheta-(-sqrt3)/(-sqrt3)=1$ or $θ = \frac{π}{4}$ $theta=pi/4$

Hence, $(- \sqrt{3}, - \sqrt{3})$ $(-sqrt3,-sqrt3)$ in polar form is $(\sqrt{6}, \frac{π}{4})$ $(sqrt6,pi/4)$

How do you convert (-sqrt(3),-sqrt(3))(−√3,−√3)(-sqrt(3),-sqrt(3)) to polar form?