What is the Cartesian form of $(-1,(14pi)/3))$ ?

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Answer 1 · 2016-10-01T03:36:43

$(x,y) = (1/2, -sqrt3/2)$

$x = r cos(theta)$

$= (-1) cos((14pi)/3)$

$= -cos((2pi)/3)$

$= -(-1/2)$

$= 1/2$

$y = r sin(theta)$

$= (-1) sin((14pi)/3)$

$= -sin((2pi)/3)$

$= -sqrt3/2$

What is the Cartesian form of (-1,(14pi)/3))(-1,(14pi)/3))?