Question

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Answer 1

\(\large \begin{array}{lllcc}
&&&x&y\\
&&rectangular&-1,& \sqrt{3}\\
&&&x&y\\
&&polar&r,& \theta\\ \quad \\ \quad \\
r^2 = x^2 + y^2 \implies r = \sqrt{x^2 + y^2}\\ \quad \\
\theta = tan^{-1}\left(tan\left(\cfrac{y}{x}\right)\right)

\end{array}\)

Answer 2

so, what do you get for "r"?

Answer 3

sq rt 10?

Answer 4

and theta 6.14?

Answer 5

hmm... I should correct that \(\bf r^2 = x^2 + y^2 \implies r = \sqrt{x^2 + y^2}\\ \quad \\
tan(\theta) = \cfrac{y}{x} \implies \theta = tan^{-1}\left(\cfrac{y}{x}\right)\)

Answer 6

same for r, theta -1.04?

Answer 7

\(\bf r = \sqrt{(-1)^2+(\sqrt{3})^2} \implies r = \sqrt{(-1)^2+(\sqrt{3^2})} \implies r = \sqrt{1+3}\)

Answer 8

r=2..

Answer 9

yeap, so r = 2, and \(\bf \theta = tan^{-1}\left(\cfrac{y}{x}\right) \implies \theta = tan^{-1}\left(\cfrac{\sqrt{3}}{-1}\right) \implies \theta = tan^{-1}\left(-\sqrt{3}\right)\)

Answer 10

if you check your Unit Circle, that angle is there :)

Answer 11

negative sq rt 3 is on the unit circle?

Answer 12

well, anyhow, lemme make it ... this way one sec

Answer 13

\(\bf tan(\theta)= \cfrac{sin(\theta)}{cos(\theta)}\\ \quad \\ \quad \\
sin(\theta) = -\cfrac{\sqrt{3}}{2}\qquad\qquad cos(\theta) = \cfrac{1}{2}\\\quad \\ \quad \\

tan(\theta) = \cfrac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} \implies -\cfrac{\sqrt{3}}{\cancel{2}} \times \cfrac{\cancel{2}}{1} \implies -\sqrt{3}\)

Answer 14

do you see now the angle where \(\bf sin(\theta) = -\cfrac{\sqrt{3}}{2}\qquad\qquad cos(\theta) = \cfrac{1}{2}\)

Answer 15

300 degrees of 5pi/3?

Answer 16

or*

Answer 17

yeap, at 5pi/3 tangent is \(\bf -\sqrt{3}\)

Answer 18

so that makes the coordinates 2, 300?

Answer 19

keep in mind that, I could have gotten also angle at \(\bf \cfrac{2\pi}{3}\) and that one also has a tangent of \(\bf -\sqrt{3}\)
however it would be OUTSIDE THE RANGE of the \(\bf tan^{-1}\) function
since the arcTangent function range is between \(\bf -\cfrac{\pi}{2}\ and \ \cfrac{\pi}{2}\)

Answer 20

yes, the polar coordinates will be \(\bf (2, 300^o) \ or \ \left(2, \cfrac{5\pi}{3}\right)\)

Answer 21

cool thanks.

Answer 22

yw