remember that while rectangular coordinates define one and only point, the position of a point using polar coordinates is not unique.

In general any point (r, Ø) can also be represented by (-r, π+Ø)

e.g. (using degrees)

look at the point defined by (6, 150°)

to get to that point, I would go 6 units in the direction of 150° or I could go (150+180)° or 330° and go in the opposite direction , then 6 units

so (6, 150°) is equivalent to (-6,330°)

in radians that would be (6, 5π/6) <-----> (-6, 11π/6)

for your (-3,4), the radius is 5, you had that

I usually get the angle in standard position by ignoring the negative sign, then using the CAST rule to place the angle in the quadrant matching the given point.

so arctan(4/3) = .9273

but (-3,4) is in II , so my angle = π - .9273 = 2.2143

which gives us the first result of (5 , 2.2143)

and our second point is (-5 , 2.2143+π)

or (-5 , 5.3559)

Of course by adding 2π or multiples of 2π , you are just adding rotations, putting you in the same spot

e.g. (5, 2.214 + 6π) would end up at the same point.

Now to your second point: (7/4 , 5/2)

or (1.75 , 2.5) since we are going to decimals anyway.

the positive r value is 3.0516 (you had that)

angle in standard position = arctan (2.5/1.75) = .9601, and our point is in I

so one answer is (3.0516, .9601)

now add π and make the r negative, to give you

(-3.0516 , .9601+π) or (-3.0516 , 4.1017)

YEAHHH ,

Since the angle is in quadrant I, the answer from your calculator was already one of the needed angles, so you just had to add π and change the sign on your r.

Adding or subtracting multiples of 2π just adds or subtracts rotations, the sign on r would not change if you work with multiples of 2π

let me know if you want me to do another example.