Homework Help: PreCalculus

Posted by Kate on Friday, January 29, 2010 at 4:25pm.

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

PreCalculus - Damon, Friday, January 29, 2010 at 4:43pm
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.
PreCalculus - Damon, Friday, January 29, 2010 at 4:48pm
By the way remember it is in quadrant 4 so negative. I just used sizes in that 30, 60, 90 triangle and ignored signs.

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