Find the rectangular coordinates for the point whose polar coordinates are given. (6√2, 5π/3)

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Find the rectangular coordinates for the point whose polar coordinates are given. (6√2, 5π/3)

Mathematics
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\[x=rcos(\theta) , y=rsin(\theta)\]
\[ \sin(5\pi/3) = \sin(\pi/3) = (\frac{\sqrt{3}}{2})\]
For (6sqrt2))cos(5pi/3) I get 4.242640607. Do I round that up or what else?

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Other answers:

what form do i put it in
\[\cos(5\pi/3) = - \cos(\pi/3) = -\frac{1}{2}\]
you can get exact values!,
they want exact values
do i not put the 6sqrt2 in the front
yeah, exact values
no, I was just finding the angles
the answers u gave are wrong.
is 1/2 the x value? and sqrt3/2 y
\[ x =-6\sqrt{2} \frac{1}{2} , y= 6\sqrt{2}\frac{\sqrt{3}}{2}\]
lol plz dont tell u you put them in :(
^ thats the FINAL answer
i did.
well simplify it a bit
x = -3sqrt(2) , y= 3sqrt(6)
the answer for x seems very obscure
the other one's r wrong to
no their not
that's what the my hw program says. i hate it :P
x = 3sqrt(2) y = -3sqrt(6)
no,
5[pi/3 is in the 4th quadrant; +x, -y
wait lols , maybe
360-60 :)
yeh I was thinking 2pi/3 :|
amistre to the rescue: Thank you for all ur help eng :)
elec did al the hard stuff; got rid of the errors I would have done lol

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