How do you convert $(-2 , 7π/6)$ to rectangular form?

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Answer 1 · 2016-11-19T21:53:25

The rectangular point is $(sqrt(3), 1)$

Given $(-2, (7pi)/6)$

Use the equations:

$x = rcos(theta) and y = rsin(theta)$

$x = (-2)cos((7pi)/6) and y = (-2)sin((7pi)/6)$

$x = (-2)(-sqrt(3)/2) and y = (-2)(-1/2)$

$x = sqrt(3) and y = 1$

The rectangular point is $(sqrt(3), 1)$

How do you convert (-2 , 7π/6) (-2 , 7π/6) to rectangular form?