Question

(-3√2, -3√2) polar coordinates?

A) 6(cos(− 4π)+i sin(−4π ))
B) 3 √2(cos(− 4π)+i sin(− 4π))
C) 6(cos( 45π )+i sin( 45π​ ))

Answer 1

Point (-3√2, -3√2) : magnitude 6, angle -450° (3rd quadrant). Polar coordinates: 6(cos(-450°) + i sin(-450°)).

The correct answer is **A)

Step-by-step explanation:

The correct answer is **A) 6(cos(-4π) + i sin(-4π))**.

Here's why:

Converting to polar coordinates:

The point in question is (-3√2, -3√2).

* The radius (magnitude) is the square root of the sum of the squares of the coordinates: √((-3√2)² + (-3√2)²) = 6.

* The angle (theta) is arctangent of the y-coordinate divided by the x-coordinate: arctangent(-3√2 / -3√2) = -4π.

Converting theta to degrees and adjusting for quadrant:

Converting theta to degrees: -4π * 180° / π = -720°.

Since the point is in the third quadrant (both x and y are negative), we need to add 270° to get the angle in the standard range for polar coordinates: -720° + 270° = -450°.

Therefore, the polar coordinates of the point (-3√2, -3√2) are 6(cos(-450°) + i sin(-450°)).

Options B and C are incorrect because:

* Option B uses the incorrect radius (3√2 instead of 6).

* Option C uses the incorrect angle (45π instead of -450°).

I hope this explanation clarifies why option A is the correct answer.