sammietaygreen
  • sammietaygreen
Solve cos(theta) ≥ sqrt(3)/2 for 0 ≤ theta ≤ 2pi
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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xapproachesinfinity
  • xapproachesinfinity
first try to solve \[\cos (\theta)=\frac{\sqrt{3}}{2}\]
xapproachesinfinity
  • xapproachesinfinity
for \[0\le \theta \le 2\pi ~~\text{what values of } \theta~~ \text{give us} \frac{\sqrt{3}}{2}\]
sammietaygreen
  • sammietaygreen
Solve for theta? I'm not sure what to do

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zepdrix
  • zepdrix
You don't know your special angles? Hmm that makes this a little more difficult..
zepdrix
  • zepdrix
|dw:1450300216980:dw|
zepdrix
  • zepdrix
Cosine corresponds to your x-coordinate.
zepdrix
  • zepdrix
So we're looking for x-coordinate of sqrt(3)/2
zepdrix
  • zepdrix
|dw:1450300253394:dw|
zepdrix
  • zepdrix
Looks like that's at these two locations, ya?
zepdrix
  • zepdrix
This is going to be a little tricky...
zepdrix
  • zepdrix
I was saying that `cosine corresponds to the x-coordinate`. Lemme draw that on the a-xis a sec. Cause we have to travel some distance in the x-direction to get to our point.
zepdrix
  • zepdrix
|dw:1450300373170:dw|
zepdrix
  • zepdrix
So here is how we get to that point, we go sqrt(3)/2 to the right, and then 1/2 up.
sammietaygreen
  • sammietaygreen
Ok I'm following so far..
zepdrix
  • zepdrix
Our problem is instructing us that this length (this cosine value) should be AT LEAST that long. It should be sqrt(3)/2 `or longer`.
zepdrix
  • zepdrix
|dw:1450300484240:dw|So if we instead went to an angle somewhere around here, do you see how this is not going to work? The x-coordinate is too short. It's below a length of sqrt(3)/2 at this location.
zepdrix
  • zepdrix
|dw:1450300561009:dw|If I look at angles over here however,
zepdrix
  • zepdrix
I made that x-coordinate length LONGER, so this is good.
zepdrix
  • zepdrix
So this is telling us that our cosine being at least sqrt(3)/2 means our angle can be anywhere... in... here
zepdrix
  • zepdrix
|dw:1450300661615:dw|
zepdrix
  • zepdrix
We have the other location though as well,
zepdrix
  • zepdrix
|dw:1450300711284:dw|
zepdrix
  • zepdrix
Again, we can make the x LONGER but not shorter.|dw:1450300745913:dw|
zepdrix
  • zepdrix
|dw:1450300779150:dw|so that means our angle can lie anywhere in here
zepdrix
  • zepdrix
So altogether, our angle theta can be anywhere in here...
zepdrix
  • zepdrix
|dw:1450300827868:dw|
zepdrix
  • zepdrix
If this doesn't make a ton of sense, that's ok. This is a very difficult Trigonometry problem, takes a lot of understanding of your functions and unit circle.
zepdrix
  • zepdrix
So now we want to write this solution as some intervals for our angle theta. And it's not just one big interval, even though it looks like it. It's a piece at the start of the circle, and a piece at the end.
zepdrix
  • zepdrix
So what's going on at the start of the circle? The circle goes FROM the angle 0, to what angle? Able to read the pink line ok?
sammietaygreen
  • sammietaygreen
Okay, so basically reading what's in between the two points?
sammietaygreen
  • sammietaygreen
like pi/6 to 0 to 11pi/6?
zepdrix
  • zepdrix
yes, but you have to read it in the correct direction. It's not the interval from pi/6 to 11pi/6. And it's also not the interval 11pi/6 to pi/6. You have to go `counter-clockwise`. So you START at angle 0, the x-axis and go counter-clockwise to angle pi/6. Ok good. That gives us half of our final answer. If \(\large\rm \cos\theta \ge\frac{\sqrt{3}}{2}\) then \(\large\rm 0\le\theta\le\frac{\pi}{6}\) That interval is part of our solution. The other interval is the other pink piece at the end of the circle.
zepdrix
  • zepdrix
So if you're going `counter-clockwise`, where would you start reading that line from?
sammietaygreen
  • sammietaygreen
it would just be reverse of what I said, right?
zepdrix
  • zepdrix
What you said didn't make a whole lot of sense I'm afraid :[ Hence, the follow up question I gave
sammietaygreen
  • sammietaygreen
I was trying to read what's in between the two dots on the pink line but I'm not sure if that's what you wanted from me
zepdrix
  • zepdrix
|dw:1450301584942:dw|
zepdrix
  • zepdrix
So we took care of this first piece by starting at angle 0 and going to angle pi/6.
zepdrix
  • zepdrix
How will you take are of the other piece that was drawn before? What angle do you start at? and where do you end up? :d
sammietaygreen
  • sammietaygreen
Wouldn't you end up back where you started at 0?
zepdrix
  • zepdrix
Yes, but you have to be careful :) It's no longer 0 if we've gone all the way around the circle. Your end point will now be 2pi instead of 0.
sammietaygreen
  • sammietaygreen
So would it be something like 0 ≤ theta ≤ pi/6, 11pi/6 ≤ theta ≤ 2pi because we went through all of those?

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