Determine all of the solutions in the interval 0< Theta < 360.
cos(Q/2) = 1/2
Q is theta

- 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

@zepdrix

- anonymous

120,360

- anikay

oh my, beaten to the punch gj

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## More answers

- lilsis76

haha okay so then hmmm

- lilsis76

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

well, now that I think about it, I think raja might be wrong

- anikay

it should have 4 solutions right?

- lilsis76

im not sure, it just says to determine all solutions in the interval 0-->360

- anonymous

yes!

- zepdrix

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

^concur

- lilsis76

lol ya, i got tired ofputting theata.

- anonymous

30

- lilsis76

okay is it 60 and 300 degrees?

- anonymous

no tetha is 30

- zepdrix

\[\large \cos(\theta/2)=\frac{1}{2}\]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

look, as I solve it, it's
\[\theta=2\cos^{-1} (1/2)\]

- anikay

the thing is, that only gives you 120. but it's a good starting point

- lilsis76

mmmmm... okay, well first. special angles is like 45-45-90, and 30-60-90, right?

- zepdrix

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

oh....okay i gotcha now on special angles. lol im totaly flipped around in this class

- lilsis76

okay then so..cosQ/2 = 1/2 right?
so cos would be a positive

- anikay

I think 120 is it.....

- zepdrix

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

yes :) 1st and 4th

- anikay

so, if we find the reference angle, then what would the two angles be?

- lilsis76

mmm....pi/3 and 5pi/6?

- anikay

yes to the first, the last, not so much, that's in the second quadrant

- anikay

or am I failing?

- zepdrix

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

oh. haha i mean 5pi/3

- anikay

With zepdrix, but you are right

- lilsis76

so the second answer would be 5pi/3?

- anikay

but, the problem is, we need that angle to be doubled to match our equation of theta/2

- lilsis76

hmmm...ok okkay, so doubled from pi/3, so pi/3 and 5pi/6?

- anikay

uh, lets throw this back to degrees shall we?
pi/3 = 60 and 5pi/3 = 300

- lilsis76

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

no, 5pi/6 = 150 degrees

- lilsis76

haha shoot what am i doing wrong

- anikay

here's a tip. every pi/6 = 30 degrees

- lilsis76

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

just plug those into your original equation and see what you get :P

- lilsis76

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

@anikay

- lilsis76

@zepdrix is it pi/3 and 2pi/3?

- zepdrix

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

yup^

- zepdrix

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

AH! so i was right?! the 60 and 300?!

- zepdrix

\[\large \theta=\quad 120^o, \quad 600^o\]

- anikay

and since we're going only to 360?

- anikay

(radians are easier imo but degrees work k)

- lilsis76

AH!!! im so lost haha

- zepdrix

aw :D

- anikay

its almost like you're my lilsis or somethin xD

- lilsis76

lol,

- lilsis76

okay so i get the 1/2 and that is in the first quad giving me 60 degrees right?

- lilsis76

@anikay

- anikay

yes, and the 300

- lilsis76

okay, how can i figure out the 300?!

- anikay

it's a reference angle of 60

- lilsis76

yes it is a ref angle 60

- lilsis76

WAIT!!! 300 cuz its cos 1.2, sin - sqrt3 /2 ?!

- lilsis76

right?! dang i think im slow

- anikay

uh

- lilsis76

haha did i do it right?!

- anikay

yes

- lilsis76

YAY!!!! > . < THANK YOU SO MUCH!!! ugh....

- anikay

however, you have to double those and since you only go to 360, 120 is your only answer

- phi

sometimes sketching the curve is helpful
|dw:1355239813847:dw|

- anikay

agreed, but my calculator is dead xD

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