Quick Polar Graph Question

I understand their working out but if you sub the values 0, pi/2, pi, 3pi/2 and 2pi as theta you get r = to 4, 4root3/3, -4, -4root3/3 and 4 respectively but the graph they've drawing hasn't pass through all those points - why?

Which points doesn't it pass through? Remember that x = rcostheta and y = rsintheta, so taking 0 for example, rcostheta = x = 4 and rsintheta = y = 0, and it passes through (4,0).

Yeah I understand that but it doesn't seem to pass through -4 when theta = pi, nor does it pass through -4root3/3 when theta equal 3pi/2

On a polar graph, a negative value of r means going the other side of the pole, O.

So r = -4 when theta = pi is the same place as r = 4 when theta = 0. In order that each point has a unique pair of coordinates, the current convention is to ignore negative values of r.

On Autograph, these points are shown as a dashed curve. Graphical calculators tend to show them in the same way as positive values of r.

Yeah I understand that but it doesn't seem to pass through -4 when theta = pi, nor does it pass through -4root3/3 when theta equal 3pi/2

If you're saying that we simply ignore all values of r which are negative then why is it when e.g. r=5 then we draw the point r=-5 to make a full circle, unless you're saying that you should draw all the values for positive rs and then reflect it?

Oh I see.

But for r^2 = a^2rootsin2theta

I found the range at which sin was positive

so that gave me 0 <= theta <= pi/2 and pi <= thea <= 3pi/2

so I made a table

theta, pi/4/ pi/2, 0, 5pi/2, 3pi/2

but when finding sin5pi/2 i got -root2/2 which is strange as I took the values where sin2theta were positive/ the midpoint of pi <= thea <= 3pi/2

Shouldn't that be 5pi/4?

Aha yeah I just noticed that after I posted it but thank you!

So am I supposed to ignore all values of theta that are negative or all values that r are negative.

Because I thought it was the former until I saw this example