How do you find the rectangular coordinates of the point with polar coordinates $(- 1, \frac{π}{3})$ $(-1,pi/3)$ ?

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Answer 1 · 2014-09-27T02:18:42

When trying to convert polar coords to rectangular coords or vice versa you need to be aware of a few equations.

$cos (θ) = \frac{x}{r} \to x = r cos (θ)$ $cos(theta)=x/r -> x=rcos(theta)$

$sin (θ) = \frac{y}{r} \to y = r sin (θ)$ $sin(theta)=y/r -> y=rsin(theta)$

$tan (θ) = \frac{y}{x} \to θ = {tan}^{- 1} (\frac{y}{x})$ $tan(theta)=y/x -> theta = tan^-1(y/x)$

$x^{2} + y^{2} = r^{2} \to r = \sqrt{x^{2} + y^{2}}$ $x^2+y^2=r^2 -> r=sqrt(x^2+y^2)$

Polar Coordinates

$(r, θ) \to (- 1, \frac{π}{3})$ $(r,theta) -> (-1,pi/3)$

Calculate the $x$ $x$ -value

$x = r cos (θ)$ $x=rcos(theta)$

$x = (- 1) \cdot cos (\frac{π}{3})$ $x=(-1)*cos(pi/3)$

$x = (- 1) (\frac{1}{2})$ $x=(-1)(1/2)$

$x = - \frac{1}{2}$ $x=-1/2$

Calculate the $y$ $y$ -value

$y = r sin (θ)$ $y=rsin(theta)$

$y = (- 1) (\frac{\sqrt{3}}{2})$ $y=(-1)(sqrt(3)/2)$

$y = - \frac{\sqrt{3}}{2}$ $y=-sqrt(3)/2$

Rectangular Coordinates

$(x, y) \to (- \frac{1}{2}, - \frac{\sqrt{3}}{2})$ $(x,y)->(-1/2,-sqrt(3)/2)$

How do you find the rectangular coordinates of the point with polar coordinates (−1,π3)(-1,pi/3)?