How do you convert $[- 6, 3 \frac{π}{2}]$ $[-6,3pi/2]$ to rectangular form?

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Answer 1 · 2016-07-10T04:46:42

Rectangular form is $(0, 6)$ $(0,6)$

Polar coordinates are given as follows: $(r, θ)$ $(r, theta)$ .

From the given problem, $r = - 6, θ = \frac{3 π}{2}$ $r=-6, theta=(3pi)/2$

We can use the following formulas to get our x and y values:
$x = r cos θ, y = r sin θ$ $x = rcostheta, y=rsintheta$

$x = r cos θ = - 6 \frac{cos (3 π)}{2} = - 6 \cdot 0 = 0$ $x = rcostheta = -6cos(3pi)/2 = -6 * 0 = 0$

$y = r sin θ = - 6 \frac{sin (3 π)}{2} = - 6 \cdot - 1 = 6$ $y = rsintheta = -6sin(3pi)/2 = -6 * -1 = 6$

Therefore, our coordinate is $(0, 6)$ $(0,6)$

How do you convert [-6,3pi/2][−6,3π2] [-6,3pi/2] to rectangular form?