For the following graph, state the polar coordinate with a positive r and positive q (in radians). Explain your steps as to how you determined the coordinate (in your own words). I'm looking for answers that involve π, not degrees for your angles. State the polar coordinate with (r, -q). Explain how you found the new angle. State the polar coordinate with (-r, q). Explain how you found the new angle. State the polar coordinate with (-r, -q). Explain how you found the new angle.

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Step-by-step explanation:

For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.

So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,

( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.

( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120

Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.

( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )

So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

sina-sinb

cosb+cosa

tana-tanb

cos0+cos x pi/3

20tan

step-by-step explanation:

where is the option given