Question

Find cot(5pi/12) using half-angle identities. So I'll take the tangent instead and then the reciprocal of that result would be my answer. So tan[(5pi/6)/2]=[1-cos(5pi/6)]/[sin(5pi/6)] = [1-(-sqrt3)/2]/[1/2] = [1+(sqrt3)/2]*2..... right... because from the previous step my textbook says that simplifies to just sqrt3. If I'm wrong I'm an idiot. Please clarify...

Answer 1

I think you've not done this correctly. What is correct is that:

tan(x/2) = sin x / (1 + cos x)

Answer 2

hence cot(x/2) = (1 + cos x)/sin x

Answer 3

And (1-cosx)/sinx right

Answer 4

and so (1+sqrt3/2)/(1/2)=2+2(sqrt3)/2 right...

Answer 5

so the tangent would be 2/(2+2(sqrt3))=1+sqrt3 ?

Answer 6

let x = 5pi/6. Then
tan(x/2) = sin x/(1 + cos x) = 1/2 / (1 - sqrt(3)/2) = 1 /(2 - sqrt(3))

Hence
cot(5pi/12) = 2 - sqrt(3)

Answer 7

Alternatively
tan(x/2)
= sin x/(1 + cos x)
= (1 - cos x)/ sin x , as you say
= (1 + sqrt(3)/2)/(1/2)
= 2 + sqrt(3)

hence

cot(5pi/2) = 1/(2 + sqrt(3))
= (2 - sqrt(3)) / [ (2 + sqrt(3)) (2 - sqrt(3)) ]
= (2 - sqrt(3)) / [ 2^2 - sqrt(3)^2 ]
= (2 - sqrt(3)) / (4 - 3)
= 2 - sqrt(3)