Answer
$\textbf{T}(t) = \langle 0, \frac{-sin(2t)}{\sqrt {1 + 3cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {1 + 3cos^2(2t)}}\rangle $
Work Step by Step
$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle 8, cos(2t), 2sin(2t)\rangle$
$\textbf{r}'(t) = \langle 0, -2sin(2t), 4cos(2t)\rangle$
$|\textbf{r}'(t)| = \sqrt {(0)^2 + (-2sin(2t))^2 + (4cos(2t))^2} = \sqrt {4sin^2(2t) + 16cos^2(2t)}$
$\textbf{T}(t) = \frac{\textbf{r}'(t)}{|\textbf{r}'(t)|} = \frac{\langle 0, -2sin(2t), 4cos(2t)\rangle}{\sqrt {4sin^2(2t) + 16cos^2(2t)}} = \langle 0, \frac{-sin(2t)}{\sqrt {sin^2(2t) + 4cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {sin^2(2t) + 4cos^2(2t)}}\rangle = \langle 0, \frac{-sin(2t)}{\sqrt {1 + 3cos^2(2t)}}, \frac{2cos(2t)}{\sqrt {1 + 3cos^2(2t)}}\rangle $
