PreCalculus
posted by Kate .
Hi I need some assistance on this problem
find the exact value do not use a calculator
cot[(5pi)/12]
Here is my attempt
RT = square root
pi = 3.14...
cot[(5pi)/12]={tan[(5pi)/12]}^1={tan[(2pi)/12(8pi)/12]}^1={tan[pi/6(2pi)/3]}^1={(tan[pi/6]tan[(2pi)/3])/(1+tan[pi/6]tan[(2pi)/3])}^1={((RT[3])/3+RT[3])/(1+((RT[3])/3)(RT3))}^1
then I got this
(RT[3] + 3RT[3])/3
over
1  3/3
as you can see this is a problem
because it leaves me with a denomenator of 0

cot (5pi/12)
well that will be in quadrant 4 since it is clockwise from the x axis due to the minus sign and 5 pi/12 is less than pi/2 or 6 pi/12.
call the x component of this vector a and the y component b. Then the magnitude of the cotan is a/b.
Notice that the angle to the negative y axis is 6pi/12  5pi/12 = pi/12
then the magnitude of tan pi/12 is a/b, just what we are looking for
so what is tan pi/12?
In degrees it is 15 degrees.
We know tan 30 degrees = 1/sqrt3 and sin 30 =1/2 and cos 30 = sqrt3 /2
so what is tan 15 degrees?
well
tan 15 = sin 30/(1+cos 30)
I think you can take it from there. 
By the way remember it is in quadrant 4 so negative. I just used sizes in that 30, 60, 90 triangle and ignored signs.