What is the Cartesian form of $( 6 , ( - 16pi)/3 )$ ?

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Answer 1 · 2017-04-10T17:25:57

$(x,y)=(-3,3sqrt(3))$

$(r, theta)=(6,-(16pi)/3)$

$x=r cos theta=6cos(-(16pi)/3)=6cdot(-1/2)=-3$

$y=r sin theta=6sin(-(16pi)/3)=6 cdot sqrt(3)/2=3sqrt(3)$

Hence, $(x,y)=(-3,3sqrt(3))$

I hope that this was clear.

What is the Cartesian form of ( 6 , ( - 16pi)/3 ) ( 6 , ( - 16pi)/3 ) ?