How do you graph the point $P (- 0.5, - \frac{11 π}{6})$ $P(-0.5, -(11pi)/6)$ ?

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How do you graph the point $P (- 0.5, - \frac{11 π}{6})$ $P(-0.5, -(11pi)/6)$ ?

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Answer 1 · 2018-07-11T15:59:54

Plot point $(- \frac{\sqrt{3}}{4}, \frac{1}{4})$ $(-\sqrt3/4, 1/4)$

Explanation:

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

$x = r cos θ = (- 0.5) cos (- \frac{11 π}{6}) = - \frac{\sqrt{3}}{4}$ $x=r\cos\theta=(-0.5)\cos(-{11\pi}/6)=-\sqrt3/4$

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

Now, plot the point $(- \frac{\sqrt{3}}{4}, \frac{1}{4})$ $(-\sqrt3/4, 1/4)$ on XY plane

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How do you graph the point $P (- 0.5, - \frac{11 π}{6})$ $P(-0.5, -(11pi)/6)$ ?

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Explanation:

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How do you graph the point $P (- 0.5, - \frac{11 π}{6})$ $P(-0.5, -(11pi)/6)$ ?