Find the rectangular coordinates for each point:

Note that for a point given in polar form `(r,theta)` the conversion to rectangular form is `(rcostheta,rsintheta)` .

(a) `(2,(3pi)/4)` ; `r=2,theta=(3pi)/4` so the rectangular form is `(2cos((3pi)/4),2sin((3pi)/4))` or `(2*(-sqrt(2))/2,2sqrt(2)/2)=(-sqrt(2),sqrt(2))`

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`(2,(3pi)/4)` in polar form is equivalent to `(-sqrt(2),sqrt(2))` in rectangular form.

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As a quick check, note that the point `(-sqrt(2),sqrt(2))` lies on the line y=-x in the second quadrant so `theta=(3pi)/4` and the length of the vector can be found using the Pythagorean theorem: `r^2=(-sqrt(2))^2+sqrt(2)^2 ==>r^2=4==>r=2` as required.

(b) `(-4,(7pi)/6)` ; The rectangular form is `(-4cos((7pi)/6),-4*sin((7pi)/6))` which simplifies to `(-4*(-sqrt(3))/2,-4*-1/2)=(2sqrt(3),2)`

The vector in polar form points into the first quadrant since r<0.

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`(-4,(7pi)/6)` in polar form is equivalent to `(2sqrt(3),2)` in rectangular form

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(c) `(2/3,-(2pi)/3)` ; The rectangular form is `(2/3cos((-2pi)/3),2/3 sin((-2pi)/3))` which simplifies to `(2/3*(-1/2),2/3*(-sqrt(3))/2)` or `(-1/3,-sqrt(3)/3)`

Since `theta=(-2pi)/3` we know the point is in the third quadrant. Again we can check that r=2/3 using the Pythagorean theorem.

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`(2/3,(-2pi)/3)` in polar form is equivalent to `(-1/3,-sqrt(3)/3)` in rectangular form

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