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Convert the rectangular coordinates to polar coordinates with r > 0 and 0 ≤ θ < 2π. (-squareroot of 6, -squareroot of 2)
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\[\theta = \arctan (-\sqrt{2}/-\sqrt{6}), r = \sqrt{6 + 2}\]
what is the radian for the \[\theta\]?
theta = .5234
is that \[\pi/2\] ?
π/2 = 1.57
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im sorry for all the questions but I want to know what the theta you got but in radians
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