anonymous
  • anonymous
What quadrant is 11pi/3 in? Find sin 0, cos 0, and tan 0. Find a positive and negative coterminal angle
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@zepdrix Can you help me on this one?
zepdrix
  • zepdrix
Well we know the boundaries for each quadrant, right? The first quadrant ends at pi/2 the second quadrant ends at pi the third quadrant ends at 3pi/2 So let's compare our 11pi/3 to each of these values and see where it lies. Get a common denominator between pi/2 and 11pi/3. Which is larger?
zepdrix
  • zepdrix
no no no let's not do that :) my bad.

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anonymous
  • anonymous
Haha okay!
zepdrix
  • zepdrix
11pi/3 is a really big angle, it's been spun around the entire circle a few times. Let's first find our co-terminal angles. We'll "unwind" it. Let's subtract some 2pi's from it.
zepdrix
  • zepdrix
\[\large\rm \frac{11\pi}{3}-2\pi=?\]
anonymous
  • anonymous
9/3?
anonymous
  • anonymous
9pi/3?
zepdrix
  • zepdrix
No. Need to get a common denominator before you can perform subtraction.
anonymous
  • anonymous
5pi/3?
zepdrix
  • zepdrix
Ok great. So that's a `positive angle` which is co-terminal with 11pi/3. How do we find a `negative angle` which is also co-terminal? Any ideas? :)
anonymous
  • anonymous
Multiply by negative 1?
anonymous
  • anonymous
It's in the second quadrant right?
zepdrix
  • zepdrix
Multiply by -1.. Hmm, no. But that was a clever idea. We'll "unwind" the angle even further. We'll subtract another 2pi from it. Q2? No.
anonymous
  • anonymous
Okay, -pi/3
anonymous
  • anonymous
What quadrant is it in?
zepdrix
  • zepdrix
Ok good :) So that answers the third question they asked.
zepdrix
  • zepdrix
-pi/3 is probably the best angle to use to figure that out. Where does -pi/3 put you?
anonymous
  • anonymous
Third?
zepdrix
  • zepdrix
Where is pi/3 located?
zepdrix
  • zepdrix
You gotta remember these three at the very least, pi/6 pi/4 and pi/3 those are important :)
anonymous
  • anonymous
Okay, im not sure where though
zepdrix
  • zepdrix
|dw:1447356525562:dw|
zepdrix
  • zepdrix
|dw:1447356583086:dw|So then if pi/3 puts us this far in the first quadrant
zepdrix
  • zepdrix
|dw:1447356628088:dw|Then negative pi/3 is the same distance but spun in the opposite direction, ya?
zepdrix
  • zepdrix
If that's too confusing, you can do the comparison thing I was talking about at the start. Let's compare 5pi/3 to 3pi/2. They need a 6 for a denominator.\[\large\rm \frac{10\pi}{6}>\frac{9\pi}{6}\]So we can see that our angle 5pi/3 is larger than the boundary for the third quadrant. So it spills over into the 4th quadrant, if not further. We can compare 5pi/3 to 2pi, and see that it's `less than` 2pi, so it's stuck in the 4th quadrant.

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