How do you find the rectangular equation for $θ = \frac{5 π}{6}$ $theta=(5pi)/6$ ?

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Answer 1 · 2016-10-01T05:17:37

$y = - \frac{x}{\sqrt{3}}$ $y=-x/sqrt 3$ , limited to the half-line in $Q_{2}$ $Q_2$ .

The conversion formula is $(r (cos θ, s i m θ) = (x, y,)$ $(r(cos theta, sim theta ) = (x, y, )$ , giving

$x = r cos θ, y = r sin θ a d e r = \sqrt{x^{2} + y^{2}} \geq 0$ $x = r cos theta, y = r sin theta ade r =sqrt(x^2+y^2)>=0$ .

Here $θ = \frac{5}{6} π$ $theta =5/6pi$ represents the half-line from pole in the

direction in $Q_{2}$ $Q_2$ .

Sans #r = 0,

y/x= sin(5/6pi)/cos(5/6pi)#

$= \frac{sin (π - \frac{π}{6})}{cos (π - \frac{π}{6})}$ $= sin(pi-pi/6)/cos(pi-pi/6)$

$= \frac{sin (\frac{π}{6})}{- cos (\frac{π}{6})}$ $=sin(pi/6)/(-cos(pi/6)$

#=-1/sqrt 3.

As a matter of fact, this half line is discontinuous at its end (0, 0).

How do you find the rectangular equation for theta=(5pi)/6θ=5π6theta=(5pi)/6?