Answer
(a) (i) $(6 , \frac {\pi} 6)$; (ii) $(-6 , \frac {7\pi} 6)$
(b) (i) $(\sqrt 5, 2\pi - tan^{-1}(2))$ (ii) $(-\sqrt 5, \pi - tan^{-1}(2))$
Work Step by Step
(a)
1. Find the angle:
$tan(\theta) = \frac{y}{x} = \frac3 {3\sqrt 3} = \frac{\sqrt 3}{3}$
$tan(\theta) = \frac {\sqrt 3} 3 \longrightarrow \theta = tan^{-1}(\frac{\sqrt 3} 3) = \frac{\pi}{6} or \frac{7\pi}{6}$
2. Calculate $r$:
$3 \sqrt 3 = rcos(\theta) = rcos(\frac{\pi}{6})$
$3 \sqrt 3 = \frac {\sqrt 3} 2 r$
$r = 6 $
$3\sqrt 3= r cos(\frac{7\pi}{6})$
$3 \sqrt 3 = r(-\frac{\sqrt 3}{2})$
$r = -6$
Thus, the points are: $(6 , \frac {\pi} 6)$ and $(-6 , \frac {7\pi} 6)$
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(b)
1. Find $r$:
$r^2 = x^2 + y^2 = 1^2 + (-2)^2 = 5$
$r = \sqrt 5$
2. Find the angle:
$tan (\theta) = \frac y x = \frac {-2} 1 = -2$
$\theta = tan^{-1} (-2) = -tan^{-1}(2)$
Since the exercise requires that $0 \leq \theta \leq 2\pi$, $ -tan^{-1}(2)$ is not valid, so we should add $2\pi$ to it, so we get the same angle, but between $0$ and $2\pi$.
$\theta = 2\pi - tan^{-1}(2)$
3. Determine the oposite angle.
To do that, we should add or remove $\pi$ from the angle. In this case, we will remove, so the angle isn't greater than $2\pi$.
$\theta_2 = 2\pi - tan^{-1}(2) - \pi = \pi - tan^{-1}(2)$
This angles corresponds to the same point, if we multiply r by $-1$.
$r = -\sqrt 5$
Thus, these are the points: $(\sqrt 5, 2\pi - tan^{-1}(2))$ and $(-\sqrt 5, \pi - tan^{-1}(2))$.
