Finding parametric equations for the tangent line at a point on a curve 
Find parametric equations for the tangent line at the point $(\cos(-\frac{4 \pi}{6}), \sin(-\frac{4 \pi}{6}), -\frac{4 \pi}{6}))$ on the curve $x = \cos(t), y = \sin(t), z=t$

I understand that in order to find the solution, I need to use partial derivatives. However, the method in my textbook works for simpler problems -- I seem to be making a calculation error when I try to apply the method to this problem.
Can anyone suggest how to approach this problem?
I found a very similar problem and solution here, but the solution by the person who answered is hard for me to follow. Unfortunately, I get stuck at the line where he subtracts $\frac{\pi}{6}$ from $\pi$ within the trigonometric functions.
Here is the "simple" method that I was originally using.
Any sincere help would be appreciated. Thank you.
 A: So $\textbf{r}(t) = \left< \cos t, \sin t, t \right>$. Then $\textbf{r}'(t) = \left<-\sin t, \cos t, 1 \right>$. So $t = -4 \pi/6$. So $\textbf{r'}(-\frac{4 \pi}{6}) = \left<-\sin( -\frac{4 \pi}{6}), \cos \left( -\frac{4 \pi}{6} \right), 1 \right>$. So the equation of the tangent line would be 
$$x = \cos(-\frac{4 \pi}{6})+t\left(-\sin\left( -\frac{4 \pi}{6}\right)\right)$$ 
$$y = \sin(-\frac{4 \pi}{6})+t\left(\cos \left(-\frac{4 \pi}{6} \right) \right)$$ 
and $$z = -\frac{4 \pi}{6}+t$$
A: I'm pretty sure that the answer to which you linked on Yahoo! Answers is just wrong.  The parametric equations $x=\cos t$, $y=\sin t$, $z=t$ describe a spiral on a cylinder of radius 1 coaxial with the $z$-axis.  Since there is only 1 parameter, these parametric equations cannot describe a 2-dimensional surface.  The method used in your second link seems appropriate—the direction vector of the tangent line at any point on $\langle x(t),y(t),z(t)\rangle=\langle\cos t,\sin t,t\rangle$ is $\langle x'(t),y'(t),z'(t)\rangle=\cdots$ (no partial derivatives needed) and you know a point on the line, so you can write a parametric equation for the tangent line.
A: the correct answer for $x$ and $y$ are:
$$
\cos(-4\pi/6) + t(-\sin(-4\pi/6))\\
\sin(-4\pi/6) + t(\cos(-4\pi/6))
$$
this is because the point that lies on the parametric lines are defined by inputting the $t$ value into the original parametric equation
A: X= cos(-4pi/6) + t(-sin(-4pi/6))
y= sin(-4pi/6) + t(cos(-4pi/6))
z= -4pi/6 + t
x= .9993 + t(0.0365)
y= -0.0365 + t(0.9993)
z= -2.094 + t
All of these values were estimated to 4 digits by plugging in (-4pi/6) into the equations for x,y,z
