lilsis76
Determine all of the solutions in the interval 0< Theta < 360.
cos(Q/2) = 1/2
Q is theta
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lilsis76
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@zepdrix
Raja99
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120,360
anikay
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oh my, beaten to the punch gj
lilsis76
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haha okay so then hmmm
lilsis76
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how would that be? @zepdrix i dont get it. i see cos Q/2 that would mean cos = 1/2 right? and if its positive it would be in the 1st and 4th Quadrant?
anikay
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well, now that I think about it, I think raja might be wrong
anikay
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it should have 4 solutions right?
lilsis76
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im not sure, it just says to determine all solutions in the interval 0-->360
Raja99
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yes!
zepdrix
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Using Q? Oooo I like that :) I'm always having trouble finding a suitable letter for theta in text, that works quite well XD
anikay
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^concur
lilsis76
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lol ya, i got tired ofputting theata.
Raja99
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30
lilsis76
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okay is it 60 and 300 degrees?
Raja99
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no tetha is 30
zepdrix
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\[\large \cos(\theta/2)=\frac{1}{2}\]We need to solve for theta, which involves a little more than just finding a special angle. If it helps, think of the ENTIRE inside of cosine as an angle, like this,\[\large \cos(\phi)=\frac{1}{2}\]Where is phi =1/2? Which special angles?
anikay
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look, as I solve it, it's
\[\theta=2\cos^{-1} (1/2)\]
anikay
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the thing is, that only gives you 120. but it's a good starting point
lilsis76
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mmmmm... okay, well first. special angles is like 45-45-90, and 30-60-90, right?
zepdrix
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The special angles are 0,30, 45, 60, 90
In radians 0, pi/6, pi/4, pi/3, pi/2 and so on...
Remember those? :)
We're not talking about special triangles :D
Special angles on the unit circle.
lilsis76
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oh....okay i gotcha now on special angles. lol im totaly flipped around in this class
lilsis76
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okay then so..cosQ/2 = 1/2 right?
so cos would be a positive
anikay
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I think 120 is it.....
zepdrix
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yes the cosine is producing a positive value, so we're concerned with angles in the 1st and 4th quadrant right? :O
Cuz that's where cosine is positive.
lilsis76
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yes :) 1st and 4th
anikay
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so, if we find the reference angle, then what would the two angles be?
lilsis76
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mmm....pi/3 and 5pi/6?
anikay
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yes to the first, the last, not so much, that's in the second quadrant
anikay
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or am I failing?
zepdrix
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lol :) They should both either be /6 or /3 :D you should get similar angles in that sense.
Also, the domain of theta was given to us in DEGREES. let's try to keep it in degrees :)
lilsis76
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oh. haha i mean 5pi/3
anikay
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With zepdrix, but you are right
lilsis76
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so the second answer would be 5pi/3?
anikay
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but, the problem is, we need that angle to be doubled to match our equation of theta/2
lilsis76
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hmmm...ok okkay, so doubled from pi/3, so pi/3 and 5pi/6?
anikay
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uh, lets throw this back to degrees shall we?
pi/3 = 60 and 5pi/3 = 300
lilsis76
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DANG! haha okay, so first we have 60 and u said to double it, so that would be 120, 120 would be.....that 5pi/6
anikay
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no, 5pi/6 = 150 degrees
lilsis76
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haha shoot what am i doing wrong
anikay
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here's a tip. every pi/6 = 30 degrees
lilsis76
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pi/6 pi/4 pi/3 pi/2 4pi/6div by 2 = 2pi/3
OH!.....haha okay so pi/3 and 2 pi/3???
anikay
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just plug those into your original equation and see what you get :P
lilsis76
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i dont have an equatioN haha says determine all of the solutions in the interval 0< Q<360
cos(Q/2) = 1/2
so the answers would be pi/3 and 2pi/3? right?
lilsis76
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@anikay
lilsis76
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@zepdrix is it pi/3 and 2pi/3?
zepdrix
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If the problem had been,\[\large \cos\left(\theta\right)=\frac{1}{2}\]We would have concluded that,\[\large \theta=\quad 60^o,\quad 300^o\]
But our problem involved Q/2, not Q.
Giving us,\[\large \cos\left(\frac{\theta}{2}\right)=\frac{1}{2}\]
\[\frac{\theta}{2}=\quad 60^o,\quad 300^o\]
From here, if we multiply both sides by 2, we can solve for Q.
anikay
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yup^
zepdrix
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I'm a little confused why you're messing with the radians :C They're a little more confusing to work with.
The domain of Q was GIVEN to us in DEGREES. So let's work in degrees! :D
lilsis76
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AH! so i was right?! the 60 and 300?!
zepdrix
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\[\large \theta=\quad 120^o, \quad 600^o\]
anikay
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and since we're going only to 360?
anikay
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(radians are easier imo but degrees work k)
lilsis76
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AH!!! im so lost haha
zepdrix
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aw :D
anikay
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its almost like you're my lilsis or somethin xD
lilsis76
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lol,
lilsis76
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okay so i get the 1/2 and that is in the first quad giving me 60 degrees right?
lilsis76
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@anikay
anikay
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yes, and the 300
lilsis76
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okay, how can i figure out the 300?!
anikay
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it's a reference angle of 60
lilsis76
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yes it is a ref angle 60
lilsis76
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WAIT!!! 300 cuz its cos 1.2, sin - sqrt3 /2 ?!
lilsis76
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right?! dang i think im slow
anikay
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uh
lilsis76
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haha did i do it right?!
anikay
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yes
lilsis76
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YAY!!!! > . < THANK YOU SO MUCH!!! ugh....
anikay
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however, you have to double those and since you only go to 360, 120 is your only answer
phi
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sometimes sketching the curve is helpful
|dw:1355239813847:dw|
anikay
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agreed, but my calculator is dead xD