How do you graph the point $P(-0.5, -(11pi)/6)$ ?

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Answer 1 · 2018-07-11T15:59:54

Plot point $(-\sqrt3/4, 1/4)$

The Cartesian coordinates $(x, y)$ of the point $(r, \theta)\equiv(-0.5, -{11\pi}/6)$ are given as follows

$x=r\cos\theta=(-0.5)\cos(-{11\pi}/6)=-\sqrt3/4$

$y=r\cos\theta=(-0.5)\sin(-{11\pi}/6)=1/4$

Now, plot the point $(-\sqrt3/4, 1/4)$ on XY plane

How do you graph the point P(-0.5, -(11pi)/6)P(-0.5, -(11pi)/6)?