Determine two pairs of polar coordinates for the point 2

Asked by maham237 @ 21/05/2021 in Mathematics viewed by 386 People

(2, -2) with 0° ≤ θ < 360°.

Answered by maham237 @ 21/05/2021

$\bf \begin{array}{rlclll}
(2&,&-2)\\
\uparrow &&\uparrow \\
x&&y
\end{array}\qquad 
\begin{cases}
r=\sqrt{x^2+y^2}\\
\theta =tan^{-1}\left( \frac{y}{x}\right)\\
----------\\
r=\sqrt{(2)^2+(-2)^2}\\
\qquad \sqrt{8}\\
\qquad 2\sqrt{2}\\
\theta =tan^{-1}\left( \frac{-2}{2} \right)\\
\qquad tan^{-1}(-1)\\\\
\qquad \cfrac{3\pi }{4}\ ,\ \boxed{\frac{7\pi }{4}}
\end{cases}\implies \left(2\sqrt{2}\ ,\ \frac{7\pi }{4} \right)$

now, notice, there are two feasible angles, they both have a tangent of -1, however, our x,y pair is +,-, the x is positive and the y is negative, that means the rectangular point is in the IV quadrant, thus, the angle is in the IV quadrant, not the II quadrant.

Determine two pairs of polar coordinates for the point 2

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