What is the Cartesian equivalent of polar coordinates $(2, \frac{π}{6})$ $(2, pi/6)$ ?

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Answer 1 · 2014-10-02T00:15:14

$(r, θ) \to (2, \frac{π}{6})$ $(r,theta)->(2,pi/6)$

$(x, y) \to (r cos (θ), r sin (θ))$ $(x,y)->(rcos(theta),rsin(theta))$

Substitute in $r$ $r$ and $θ$ $theta$

$(x, y) \to (2 cos (\frac{π}{6}), 2 sin (\frac{π}{6}))$ $(x,y)->(2cos(pi/6),2sin(pi/6))$

Remember back to the unit circle and special triangles.

$\frac{π}{6} = 30^{\circ}$ $pi/6=30^circ$

$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos(pi/6)=sqrt(3)/2$

$sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$

Substitute in those values.

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

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

What is the Cartesian equivalent of polar coordinates (2,π6)(2, pi/6)?