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Trigonometry 12 Online
OpenStudy (anonymous):

If cosA = 1/2 with A in QIV find cos A/2

OpenStudy (fibonaccichick666):

First you need to figure out what A is, any ideas?

OpenStudy (anonymous):

Would it be 300 degrees?

OpenStudy (fibonaccichick666):

Explain how you arrived at that conclusion for me please

OpenStudy (anonymous):

If cos of A is 1/2 and in quadrant 4 on the unit circle shows 300... Am I off on thinking that way?

OpenStudy (fibonaccichick666):

uhm, well let me show you how I think about it. (it is not 300) So in the 1st quad, which angle has a cosine of 1/2?

OpenStudy (anonymous):

60

OpenStudy (fibonaccichick666):

I apologize, I was incorrect. It is 300, but let's still say why

OpenStudy (fibonaccichick666):

so if we graph,|dw:1395190978468:dw|

OpenStudy (fibonaccichick666):

Can you tell me the value of each side in terms of x?

OpenStudy (fibonaccichick666):

and label however you choose for the angles

OpenStudy (anonymous):

Not sure I follow

OpenStudy (fibonaccichick666):

so you know how for 45 deg we get a 1, 1, √2 triangle, how about for the one with 60 deg

OpenStudy (anonymous):

x, 2x, x sqrt 3

OpenStudy (fibonaccichick666):

good so can you label those on our picture?

OpenStudy (anonymous):

x= x, h= 2x, y= x sqrt 3

OpenStudy (fibonaccichick666):

|dw:1395192172888:dw| And what would be the value of angle z?

OpenStudy (anonymous):

300

OpenStudy (fibonaccichick666):

in regards to the y axis

OpenStudy (anonymous):

30

OpenStudy (fibonaccichick666):

ok so that is how we are going to draw A/2. You have determined that A is 300 now A/2=?

OpenStudy (anonymous):

150

OpenStudy (fibonaccichick666):

so can you graph that for me please

OpenStudy (anonymous):

I'm on my iPad and it can't access the utilities

OpenStudy (fibonaccichick666):

k draw it for yourself then

OpenStudy (fibonaccichick666):

then tell me which quadrant it falls in and what angle is formed with the y axis

OpenStudy (anonymous):

QII and 60 degrees

OpenStudy (fibonaccichick666):

|dw:1395193203972:dw|

OpenStudy (fibonaccichick666):

good, so now, can you figure out what the cosine value is?

OpenStudy (anonymous):

-sqrt. 3/2 ?

OpenStudy (fibonaccichick666):

yuppers, and why?

OpenStudy (anonymous):

Idk because if I use the half angle formula I get sqrt of (3/4)...

OpenStudy (fibonaccichick666):

well... you could use that, but it's a lot easier to just look at the pic. The reference angle will always be the one that is created by the original angle and the x axis

OpenStudy (fibonaccichick666):

|dw:1395194285467:dw| Think about how we labelled the sides before

OpenStudy (fibonaccichick666):

|dw:1395194307787:dw|

OpenStudy (fibonaccichick666):

use that to come to your conclusion for this new triangle

OpenStudy (anonymous):

Well my prof wants to see that formula.. Is that answer I got right?

OpenStudy (fibonaccichick666):

oh well, i didn't know that haha but tell me it again? you gave me two answers. Also you should learn this way it's a heck of a lot easier than remembering all of the formulas

OpenStudy (anonymous):

Haha no worries. I got -sqrt of 3/4

OpenStudy (fibonaccichick666):

yup that's it

OpenStudy (fibonaccichick666):

I personally do it this way. I can't remember my name half the time so this def helps

OpenStudy (anonymous):

Ha right on, thank you!

OpenStudy (fibonaccichick666):

np have a good one

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