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)#