Difficult calculus question

[JackSonSmith]

A curve has parametric equations  x= t - cos(t)  and y = sin(t)
Find the equation to the curve when t = pi/6

I keep on getting y = (root 3 /3) x  -  (root 3) pi / 18  + 1/2

However the answers say that it's y = (root 3 /3) x  -  (root 3) pi / 18  + 1

[kinslayer]

A curve has parametric equations  x= t - cos(t)  and y = sin(t)
Find the equation to the curve when t = pi/6

I keep on getting y = (root 3 /3) x  -  (root 3) pi / 18  + 1/2

However the answers say that it's y = (root 3 /3) x  -  (root 3) pi / 18  + 1

r'(t) = (x'(t), y'(t)) = (1 + sin(t), cos(t))

r'(pi/6) = (3/2, sqrt(3)/2)

A tangent to the curve is r(pi/6) + t*r'(pi/6) = (pi/6 - sqrt(3/2), 1/2) + (3*t/2, sqrt(3)*t/2)

= (pi/6 - sqrt(3)/2 + 3*t/2, 1/2 + sqrt(3)*t/2)

Converting to cartesian we have

t = (2/3)*(x + sqrt(3)/2 - pi/6) = (y - 1/2)*(2/sqrt(3))

after algebra we get :

y = sqrt(3)*x/3 - sqrt(3)*pi/18 + 1