cos(9pi/2-pi/6)

Hi, I can see that you are a new member here at OpenStudy so, I would first like to say Welcome to OpenStudy, I would like to point you to the chat pods these chat pods are where you can make new friends and talk to new people just like you, I would also like to emphasize our "NON-CHEATING POLICY" we ask here at OpenStudy that you do NOT post exam/Test questions if you are caught doing this we will notify your current school I would also like to ask you to check out the OpenStudy "Code of Conduct" http://openstudy.com/code-of-conduct Please read this carefully and thoroughly. Welcome to the OpenStudy Community! Ambassador

Knowing cos(pi/3) = 1/2. We let the denominators in your function be some pi/6. 27pi/6 - pi/6 = 26pi/6, which simplifies to 13pi/3. Since pi/3 = 1/2. 13pi/3 = 13/2.

13/2 is not correct :(

Ahh how silly of me, 1/2. I'm forget this is a unit circle.

i kn01 cos(pi/6)=\[\sqrt{3}\]/2 and sin(pi/6)=1/2

ok let me try that (:

1/2 is correct! can you explain how you knew that?

Yes, 13pi/3 would be right. Unit circle in radians ranges from (0, 2pi) 3 goes into 13 4 times with pi/3 remaining which = 1/2

It uses some trig and pre-calculus knowledge. http://en.wikipedia.org/wiki/Unit_circle

this is homework for my precal class and its the first time i have ever taken precal and is ridiculously hard!

Join our real-time social learning platform and learn together with your friends!