Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

How do you find the value of cot 14pi/3 Find the Value using Unit Circle

OpenStudy (anonymous):

you'll notice from the unit circle that the values of cot repeat on a cycle every pi/2 radians. this is because it is a periodic function with a period of pi/2. Looking at a graph of cotan, by googling it, might help visualize. So: \[\cot (x) = \cot(x \pm k\frac{ \pi }{ 2 })\]Where k is any integerSo:\[\cot(\frac{ 14 \pi }{ 3 }) = \cot( \frac{ 14\pi }{ 3 } \pm k \frac{ \pi }{ 2 })\] keep plugging in k values (1,2,3,4, ect..) until you fall between 0 and 2 pi and you can use the value using the unit circle

OpenStudy (anonymous):

I understand 14pi/3 is 840 Degrees but I need the answer in radians. I can do the trig functions with sin cos and tan, but csc sec and cot is where I am not understanding what to do. Cot = 1/tan so does that mean 1/ 14pi/3?

OpenStudy (anonymous):

no. 1/tan(14pi/3) = cot(14pi/3)

OpenStudy (anonymous):

you can also use cot(x) = cos(x)/sin(x)

OpenStudy (anonymous):

tab 14pi/3 = -\[\sqrt{3}\] so does that mean cot 14pi/3 = - 1 / \[\sqrt{3}\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

How do you your answer when you get something such as \[2\sqrt{3}\div3\]

OpenStudy (anonymous):

when you have to put it over 1

OpenStudy (anonymous):

you can't really simplify it. I'd present it as: \[\frac{ 2\sqrt3 }{ 3 }\] unless you mean, to submit to an online quiz or whatever

OpenStudy (anonymous):

Takehome quiz, but another question Division by 0 is undefined correct?

OpenStudy (anonymous):

correct

OpenStudy (anonymous):

Thanks for your help!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

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