Sunday
April 19, 2015

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

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