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Determine two pairs of polar coordinates for the point (4, -4) with 0° ≤ θ < 360°
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x=4,y=-4 let \[4=r~\cos \theta,and ~-4=r \sin \theta \] square and add \[r^2(\cos ^2\theta+\sin ^2\theta )=4^2+(-4)^2=32\] \[r=\pm \sqrt{32}=\pm 4\sqrt{2}\] \[\sin \theta ~is~ negative ~and \cos \theta~is~positive.\] so theta lies in 4th quadrant. when \[r=4\sqrt{2},\cos \theta=\frac{ 4 }{ 4\sqrt{2} }=\frac{ 1 }{ \sqrt{2} }=\cos \left( 2 \pi-\frac{ \pi }{ 6 } \right)\] \[\theta=\frac{ 11 \pi }{ 6 }\] so polar co-ordinates are \[(4 \sqrt{2},\frac{ 11 \pi }{ 6 })\] when \[r=-4\sqrt{2 },find \theta\]
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