Help me start this problem?

[helpmeplease40]

To start this i think i need to make a reference triangle first but im not sure what quadrant it would be in. After I get that info i think i can do the rest myself.

 

[Steven G]

Yes you need to draw a reference angle.

I will write t instead of theta.

You are told that that 5pi/2 < t <3pi. Both 5pi/2 = 2 1/2 pi and 3 pi are larger than 2 pi. You can always add to subtract 2pi to an angle without changing the location of the angle. A big hint is to subtract 2pi from both 5pi/2 and 3pi. This will give you an angle between 0pi and 2pi.

Of course, if you prefer, you can change the given angles into degrees and do exactly what I suggested above.

Please give it a try and post back with your work including a diagram.

 

[Deleted member 4993]

5π/2 < Θ < 3π/2 .... →.... (π + π/4) < Θ/2 < (π+ π/2)

In which quadrant does Θ/2 lie → 3rd. quadrant. Thus take care of ± accordingly.

 

[Steven G]

5π/2 < Θ < 3π/2 .... →.... (π + π/4) < Θ/2 < (π+ π/2)

In which quadrant does Θ/2 lie → 3rd. quadrant. Thus take care of ± accordingly.

To OP: Subhotosh Khan method is cleaner than what I wrote.